- First find the distance from the center to the given point. - A sphere consists of all points a fixed distance from its center. - Square the radius in the coordinate equation.

1. The radius is the distance from \(M\) to \(P\): \(r=\sqrt{(7-3)^2+(2-(-1))^2+(4-4)^2}=5\). 2. In vector form, the sphere is \(\lVert\mathbf{x}-\langle 3,-1,4\rangle\rVert=5\). 3. In coordinate form, the sphere is \((x-3)^2+(y+1)^2+(z-4)^2=25\).

Vector form: \(\lVert\mathbf{x}-\langle 3,-1,4\rangle\rVert=5\) Coordinate form: \((x-3)^2+(y+1)^2+(z-4)^2=25\)

- Read the center and radius from the equation. - Compare each point's squared distance from the center with \(r^2\). - You do not need to take square roots.

1. The center is \(M(1,-4,2)\), and \(r^2=49\). 2. For \(A\), the squared distance to \(M\) is \(0^2+6^2+(-3)^2=45\). Since \(45<49\), \(A\) is inside. 3. For \(B\), the squared distance is \(6^2+(-2)^2+3^2=49\). Therefore, \(B\) is on the sphere. 4. For \(C\), the squared distance is \((-4)^2+4^2+6^2=68\). Since \(68>49\), \(C\) is outside.

\(A\) is inside; \(B\) is on the sphere; \(C\) is outside.

- Identify the sphere's center from standard form. - The radius to the point of tangency is perpendicular to the tangent plane. - Use the center and tangent point to find a normal vector. - Substitute the tangent point into the plane equation.

1. The sphere's center is \(M=(3, -1, 2)\). 2. The radius vector to the point of tangency is \(\overrightarrow{MB}=\langle2, 3, 6\rangle\). This vector is normal to the tangent plane. 3. A plane with this normal vector has the form \(2x+3y+6z=d\). 4. Substitute \(B=(5, 2, 8)\): \(d=2(5)+3(2)+6(8)=64\). 5. Therefore, the tangent plane is \(2x+3y+6z=64\).

\(2x+3y+6z=64\)

- The radius to a tangent point is perpendicular to the tangent plane. - There is one possible center on each side of the plane. - Use the standard sphere equation.

1. The center must lie on the line through \(B\) perpendicular to the xy-plane. 2. Because the radius is \(3\), the center is \(3\) units above or below the plane. Thus, the possible centers are \(M_1(2,1,3)\) and \(M_2(2,1,-3)\). 3. Using standard form, the two sphere equations are \((x-2)^2+(y-1)^2+(z-3)^2=9\) and \((x-2)^2+(y-1)^2+(z+3)^2=9\).

\((x-2)^2+(y-1)^2+(z-3)^2=9\) and \((x-2)^2+(y-1)^2+(z+3)^2=9\)

- The center is the midpoint of the diameter. - The radius is half the diameter length. - Substitute \(P\) into the sphere equation.

1. The center is the midpoint of \(\overline{AB}\): \(M=\left(\frac{1+7}{2},\frac{1+9}{2},\frac{1+1}{2}\right)=(4,5,1)\). 2. The squared radius is \(MA^2=(4-1)^2+(5-1)^2+(1-1)^2=25\), so \(r=5\). 3. Vector form is \(\lVert\mathbf{x}-\langle 4,5,1\rangle\rVert^2=25\). 4. Coordinate form is \((x-4)^2+(y-5)^2+(z-1)^2=25\). 5. For \(P\), \((8-4)^2+(2-5)^2+(1-1)^2=25\), so \(P\) lies on the sphere.

a) Vector form: \(\lVert\mathbf{x}-\langle 4,5,1\rangle\rVert^2=25\); coordinate form: \((x-4)^2+(y-5)^2+(z-1)^2=25\) b) Yes, \(P\) lies on the sphere.

- A fixed distance from one point defines a sphere. - Substitute each test point into the equation. - In part c), replace the fixed coordinates before solving for y.

1. The distance condition is \((x-2)^2+(y+3)^2+(z-5)^2=49\). 2. For \(P(9,-3,5)\), the left side is \(7^2=49\), so \(P\) satisfies the condition. 3. For \(Q(6,0,5)\), the left side is \(4^2+3^2=25\), so \(Q\) does not satisfy it. 4. In part c, set \(x=2\) and \(z=5\). Then \((y+3)^2=49\), so \(y=4\) or \(y=-10\). 5. The required points are \((2,4,5)\) and \((2,-10,5)\).

a) \((x-2)^2+(y+3)^2+(z-5)^2=49\) b) \(P\) satisfies the condition; \(Q\) does not. c) \((2,4,5)\) and \((2,-10,5)\)

- Every point on a sphere is one radius from its center. - Use the three-dimensional distance formula. - A squared equation can produce two values.

1. Because \(P\) lies on the sphere, its distance from \(M\) is \(9\). 2. Set up the distance equation: \((x-2)^2+(5-1)^2+(12-4)^2=9^2\). 3. Simplify: \((x-2)^2+16+64=81\), so \((x-2)^2=1\). 4. Therefore, \(x-2=1\) or \(x-2=-1\), giving \(x=3\) or \(x=1\).

\(x=1\) or \(x=3\)

- Read the sphere's center and radius from standard form. - Compare the center-to-plane distance with the sphere's radius for tangency. - For a circular cross-section, use a right triangle formed by the sphere radius, the cross-section radius, and the center-to-plane distance. - Use the plane's normal vector to calculate the perpendicular distance.

1. The sphere has center \(M=(2, -1, 4)\) and radius \(5\). The planes have normal vector \(\mathbf{n}=\langle2, -2, 1\rangle\), whose magnitude is \(3\). 2. The perpendicular distance from \(M\) to \(E_k\) is \(d=\frac{|2(2)-2(-1)+4-k|}{3}=\frac{|10-k|}{3}\). 3. For tangency, \(d=5\). Thus, \(|10-k|=15\), giving \(k=-5\) or \(k=25\). 4. For a cross-sectional circle of radius \(4\), the sphere radius, plane distance, and circle radius satisfy \(d^2+4^2=5^2\). Hence, \(d=3\). 5. Therefore, \(\frac{|10-k|}{3}=3\), so \(|10-k|=9\), giving \(k=1\) or \(k=19\).

a) \(k=-5\) or \(k=25\) b) \(k=1\) or \(k=19\)

- Complete the square in the x- and y-terms. - Compare the point's squared distance from the center with \(r^2\). - A larger squared distance means the point is outside.

1. Complete the square: \(x^2+6x=(x+3)^2-9\) and \(y^2-4y=(y-2)^2-4\). 2. The equation becomes \((x+3)^2+(y-2)^2+z^2=25\). 3. Therefore, the center is \(M(-3,2,0)\) and the radius is \(5\). 4. For \(A\), the squared distance from the center is \((1+3)^2+(0-2)^2+(4-0)^2=36\). 5. Since \(36>25\), point \(A\) lies outside the sphere.

a) Center \(M(-3,2,0)\); radius \(5\) b) \(A\) lies outside the sphere.

- Complete the square in all three variables. - Compare the center-to-point distance with the radius. - For an exterior point, subtract the radius from the center-to-point distance.

1. Complete the square to obtain \((x+3)^2+(y-2)^2+(z+1)^2=25\). 2. The center is \(M(-3,2,-1)\), and the radius is \(5\). 3. The distance from \(P\) to \(M\) is \(\sqrt{(9+3)^2+(7-2)^2+(-1+1)^2}=\sqrt{169}=13\). 4. Because \(13>5\), \(P\) is outside the sphere. 5. The shortest distance to the surface is \(13-5=8\).

a) Center \(M(-3,2,-1)\); radius \(5\) b) \(13\) units c) \(P\) is outside; its distance to the surface is \(8\) units.

- Read the center and radius from the sphere equation. - Use the plane's normal vector to calculate the perpendicular distance from the center. - Compare that distance with the sphere's radius. - Use the right triangle formed by the sphere radius, the plane distance, and the cross-section radius.

1. The sphere has center \(M=(2, -3, 1)\) and radius \(R=10\). 2. The plane has normal vector \(\mathbf{n}=\langle6, -3, 2\rangle\), whose magnitude is \(7\). The distance from \(M\) to the plane is \(d=\frac{|6(2)-3(-3)+2(1)+15|}{7}=\frac{38}{7}\). 3. Since \(\frac{38}{7}<10\), the plane intersects the sphere in a circle. 4. The cross-section radius is \(\rho=\sqrt{R^2-d^2}=\sqrt{100-\left(\frac{38}{7}\right)^2}=\frac{24\sqrt6}{7}\).

The plane intersects the sphere in a circle with radius \(\rho=\frac{24\sqrt6}{7}\).

- Complete the square to identify the sphere's center and radius. - Find a vector perpendicular to both direction vectors of the plane. - Compare the center-to-plane distance with the sphere's radius. - When the plane misses the sphere, subtract the radius from the center-to-plane distance.

1. Complete the square: \((x-1)^2+(y-2)^2+(z+2)^2=16\). Thus, the sphere has center \(M=(1, 2, -2)\) and radius \(R=4\). 2. A normal vector to the plane must be perpendicular to both direction vectors. The vector \(\mathbf{n}=\langle0, -3, 4\rangle\) satisfies \(\mathbf{n}\cdot\langle1, 0, 0\rangle=0\) and \(\mathbf{n}\cdot\langle0, 4, 3\rangle=0\). 3. Using the point \((5, 0, 6)\), the plane equation is \(-3y+4z-24=0\). 4. The distance from \(M\) to the plane is \(d=\frac{|-3(2)+4(-2)-24|}{5}=\frac{38}{5}\). 5. Since \(\frac{38}{5}>4\), the plane does not intersect the sphere. The shortest distance from the sphere's surface to the plane is \(d-R=\frac{38}{5}-4=\frac{18}{5}=3.6\).

The plane does not intersect the sphere. The shortest distance from the sphere's surface to the plane is \(\frac{18}{5}=3.6\).

- Complete the square to find the sphere's center. - The radius from the center to the tangent point is normal to the tangent plane. - A normal vector may be replaced by a nonzero scalar multiple. - Substitute the tangent point to determine the constant in the plane equation.

1. Complete the square: \((x+5)^2+(y-4)^2+(z+1)^2=144\). Thus, the sphere's center is \(M=(-5, 4, -1)\). 2. The radius vector to the point of tangency is \(\overrightarrow{MB}=\langle8, 8, 4\rangle\), which is parallel to \(\langle2, 2, 1\rangle\). 3. This radius vector is normal to the tangent plane, so write the plane as \(2x+2y+z=d\). 4. Substitute \(B=(3, 12, 3)\): \(d=2(3)+2(12)+3=33\). 5. Therefore, one equation of the tangent plane is \(2x+2y+z=33\).

\(2x+2y+z=33\)

- Complete the square in each variable. - The right side of the standard sphere equation is \(r^2\). - Use the plane's normal vector to find the perpendicular distance from the center. - Compare that distance with the sphere's radius.

1. Complete the square: \((x-2)^2+(y+3)^2+(z-1)^2=25\). Therefore, \(M=(2, -3, 1)\) and \(r=5\). 2. The plane has normal vector \(\mathbf{n}=\langle2, -2, 1\rangle\), whose magnitude is \(3\). 3. The distance from \(M\) to the plane is \(d=\frac{|2(2)-2(-3)+1-25|}{3}=\frac{14}{3}\). 4. Since \(\frac{14}{3}<5\), the plane intersects the sphere in a circle.

a) \(M=(2, -3, 1)\) and \(r=5\) b) \(d=\frac{14}{3}\). Since \(d<r\), the plane intersects the sphere.

- A plane is tangent when its distance from the sphere's center equals the radius. - The radius to the point of tangency is perpendicular to the plane. - For a plane parallel to the xy-plane, only the z-coordinate determines the distance. - Identify the z-values covered by the sphere.

1. The distance from \(M=(4, 0, -2)\) to the plane \(x=10\) is \(|4-10|=6\), equal to the sphere's radius. Therefore, the plane is tangent. 2. The radius to the tangent point is parallel to the x-axis and extends from \(x=4\) to \(x=10\), so \(B=(10, 0, -2)\). 3. The distance from \(M\) to the plane \(z=k\) is \(|k+2|\). The plane misses the sphere when this distance is greater than \(6\). 4. Solve \(|k+2|>6\): \(k<-8\) or \(k>4\).

a) The center-to-plane distance is \(6\), so \(E\) is tangent at \(B=(10, 0, -2)\). b) \(k<-8\) or \(k>4\)

- Substitute the point into the sphere equation. - The radius vector to the tangent point is normal to the tangent plane. - Parallel planes have parallel normal vectors. - Use the point-to-plane distance and set it equal to \(10\).

1. Substitute \(P\) into the sphere equation: \((2-4)^2+(2+2)^2+(-3-1)^2=4+16+16=36\). Thus, \(P\) lies on the sphere. 2. The center is \(M=(4, -2, 1)\). The vector \(\overrightarrow{MP}=\langle-2, 4, -4\rangle\) is normal to the tangent plane and is parallel to \(\langle1, -2, 2\rangle\). 3. A plane through \(P\) with this normal vector is \(T:x-2y+2z+8=0\). 4. Every plane parallel to \(T\) has the form \(x-2y+2z+k=0\). Its distance from \(M\) is \(\frac{|4-2(-2)+2(1)+k|}{3}=\frac{|10+k|}{3}\). 5. Set this distance equal to \(10\): \(|10+k|=30\). Thus, \(k=20\) or \(k=-40\). 6. The planes are \(F_1:x-2y+2z+20=0\) and \(F_2:x-2y+2z-40=0\).

a) Substitution gives \(36=36\), so \(P\) lies on the sphere. b) \(T:x-2y+2z+8=0\) c) \(F_1:x-2y+2z+20=0\) and \(F_2:x-2y+2z-40=0\)

- Compare the center-to-plane distance with the sphere's radius. - Use the plane's normal vector to calculate the perpendicular distance. - The sphere radius, plane distance, and cross-section radius form a right triangle. - The circle's center is the perpendicular projection of the sphere's center onto the plane.

1. The plane has normal vector \(\mathbf{n}=\langle2, -1, 2\rangle\), whose magnitude is \(3\). The distance from \(M\) to the plane is \(d=\frac{|2(1)-2+2(3)+12|}{3}=6\). 2. Since \(6<7\), the plane intersects the sphere in a circle. 3. The cross-section radius is \(r_s=\sqrt{7^2-6^2}=\sqrt{13}\). 4. The circle's center is the perpendicular projection of \(M\) onto the plane. Write \(M_s=(1, 2, 3)+t\langle2, -1, 2\rangle\). 5. Substitution into the plane equation gives \(18+9t=0\), so \(t=-2\). Therefore, \(M_s=(-3, 4, -1)\).

a) The plane intersects the sphere in a circle. b) \(r_s=\sqrt{13}\) and \(M_s=(-3, 4, -1)\)

- How does the distance from the sphere's center to the cutting plane affect the size of the circle? - Which lengths in the sphere and circular section meet at a right angle? - Why must the center of the section be the point of the plane closest to the sphere's center?

1. The sphere has center \(M=(5, -2, 3)\) and radius \(R=15\). 2. The plane's normal vector is \(\mathbf{n}=\langle6, -3, 2\rangle\), whose magnitude is \(7\). The distance from \(M\) to the plane is \(d=\frac{|6(5)-3(-2)+2(3)+7|}{7}=7\). 3. Since \(7<15\), the plane intersects the sphere in a circle. 4. The circle's radius is \(r=\sqrt{15^2-7^2}=\sqrt{176}=4\sqrt{11}\). 5. Its center is the perpendicular projection of \(M\) onto the plane. Write \(M_S=M+t\mathbf{n}\). Substitution into the plane equation gives \(49+49t=0\), so \(t=-1\). 6. Therefore, \(M_S=(5, -2, 3)-\langle6, -3, 2\rangle=(-1, 1, 1)\).

The plane intersects the sphere because \(7<15\). \(M_S=(-1, 1, 1)\) \(r=4\sqrt{11}\)

- What distinguishes a tangent line from a line that cuts through a sphere? - Which point of the line is closest to the center of the sphere? - How can you verify that your proposed contact point has the needed property?

1. A direction vector for \(g\) is \(\mathbf{u}=\langle2, 1, 1\rangle\), so \(B=(4, 1, 1)+t\langle2, 1, 1\rangle\). 2. At the point of tangency, \(\overrightarrow{MB}\) is perpendicular to \(\mathbf{u}\). Since \(\overrightarrow{MB}=\langle2+2t, -4+t, 2+t\rangle\), set \(\overrightarrow{MB}\cdot\mathbf{u}=0\). 3. This gives \(2(2+2t)+(-4+t)+(2+t)=6t+2=0\), so \(t=-\frac13\). 4. Therefore, \(B=\left(\frac{10}{3}, \frac{2}{3}, \frac{2}{3}\right)\). 5. The radius is \(MB=\sqrt{\left(\frac43\right)^2+\left(-\frac{13}{3}\right)^2+\left(\frac53\right)^2}=\frac{\sqrt{210}}{3}\).

The radius is \(r=\frac{\sqrt{210}}{3}\), and the point of tangency is \(B=\left(\frac{10}{3}, \frac{2}{3}, \frac{2}{3}\right)\).

- At which point of a line would a sphere centered at the given point first touch it? - What makes that contact different from a line passing through the sphere? - Check your radius against the geometry of the center and line.

1. Write the point of tangency as \(S=(3+s, 2s, 1-s)\). Then \(\overrightarrow{MS}=\langle2+s, 2s-1, -s-3\rangle\). 2. At the point of tangency, \(\overrightarrow{MS}\) is perpendicular to the line's direction vector \(\mathbf{v}=\langle1, 2, -1\rangle\). 3. Set the dot product equal to zero: \((2+s)+2(2s-1)-(-s-3)=6s+3=0\). Thus, \(s=-\frac12\). 4. Substitution gives \(S=\left(\frac52, -1, \frac32\right)\). 5. The radius is \(MS=\sqrt{\left(\frac32\right)^2+(-2)^2+\left(-\frac52\right)^2}=\frac{5\sqrt2}{2}\).

The radius is \(r=\frac{5\sqrt2}{2}\), and the point of tangency is \(S=\left(\frac52, -1, \frac32\right)\).

- A tangent coordinate plane is one radius from the center. - Each of the x- and y-coordinates can have either sign. - The z-coordinate is fixed by the given plane.

1. Tangency to the xz-plane \(y=0\) requires the center's y-coordinate to satisfy \(\lvert y_M\rvert=5\). 2. Tangency to the yz-plane \(x=0\) requires \(\lvert x_M\rvert=5\). 3. The center lies in \(z=10\), so \(z_M=10\). 4. The four possible centers are \((5,5,10)\), \((5,-5,10)\), \((-5,5,10)\), and \((-5,-5,10)\). 5. Substitute each center and \(r^2=25\) into the standard sphere equation.

\((x-5)^2+(y-5)^2+(z-10)^2=25\) \((x-5)^2+(y+5)^2+(z-10)^2=25\) \((x+5)^2+(y-5)^2+(z-10)^2=25\) \((x+5)^2+(y+5)^2+(z-10)^2=25\)

- Compare the squared distance from \(P\) to the center with \(r^2\). - Solve the equality first to find the boundary values. - Use the corresponding quadratic inequality for the inside and outside cases.

1. The squared distance from \(P\) to \(M\) is \(d^2=(4-1)^2+(1-1)^2+(c-1)^2=9+(c-1)^2\). 2. On the sphere, \(9+(c-1)^2=25\), so \((c-1)^2=16\). Thus, \(c=-3\) or \(c=5\). 3. Inside the sphere, \((c-1)^2<16\), which gives \(-3<c<5\). 4. Outside the sphere, \((c-1)^2>16\), which gives \(c<-3\) or \(c>5\).

Inside: \(-3<c<5\) On: \(c=-3\) or \(c=5\) Outside: \(c<-3\) or \(c>5\)

- Write the squared distance as a quadratic expression in \(c\). - Solve the equality to find the boundary values. - Use the sign of the upward-opening quadratic between and outside its roots.

1. The squared distance is \(d^2=c^2+(c-1)^2+1=2c^2-2c+2\). 2. On the sphere, \(2c^2-2c+2=6\). This simplifies to \(c^2-c-2=0\), so \(c=-1\) or \(c=2\). 3. Because \(2c^2-2c-4\) opens upward, \(d^2<6\) between the roots: \(-1<c<2\). 4. Outside the sphere, \(c<-1\) or \(c>2\).

Inside: \(-1<c<2\) On: \(c=-1\) or \(c=2\) Outside: \(c<-1\) or \(c>2\)

- Write the sphere equation from its center and radius. - Substitute the parametric coordinates of the line into the sphere equation. - The number of real parameter values determines the number of intersections. - Substitute each parameter value back into the line equation.

1. The sphere equation is \((x-2)^2+y^2+(z-7)^2=9\). 2. Substitute \(x=1+t\), \(y=2-2t\), and \(z=5+2t\): \((t-1)^2+(2-2t)^2+(2t-2)^2=9\). 3. Simplifying gives \(9(t-1)^2=9\), so \((t-1)^2=1\). Therefore, \(t=0\) or \(t=2\). 4. For \(t=0\), the point is \((1, 2, 5)\). For \(t=2\), the point is \((3, -2, 9)\). 5. Because there are two intersection points, \(g\) is a secant line of the sphere.

Line \(g\) is a secant. The intersection points are \((1, 2, 5)\) and \((3, -2, 9)\).

- Substitute the line's coordinate expressions into the sphere equation. - Expand the squared binomials carefully. - Use the number of real solutions of the resulting quadratic equation. - One real parameter value means the line is tangent.

1. Substitute \(x=6+2\lambda\), \(y=-\lambda\), and \(z=5\) into the sphere equation: \((2\lambda+3)^2+(1-\lambda)^2+3^2=14\). 2. Expanding and simplifying gives \(5\lambda^2+10\lambda+5=0\). 3. Divide by \(5\): \((\lambda+1)^2=0\). Thus, there is exactly one parameter value, \(\lambda=-1\), so the line is tangent to the sphere. 4. Substitution into \(h\) gives the point of tangency \((4, 1, 5)\).

Line \(h\) is tangent to the sphere at \((4, 1, 5)\).

- Parallel planes have parallel normal vectors. - A tangent plane is one radius from the sphere's center. - Move from the center in both unit-normal directions to locate the tangent points. - Substitute each tangent point into a plane equation with the given normal vector.

1. The sphere has center \(M=(2, 0, -1)\) and radius \(6\). The given plane has normal vector \(\mathbf{n}=\langle2, 1, -2\rangle\), whose magnitude is \(3\). 2. The tangent points lie \(6\) units from \(M\) in the two normal directions. Since \(\frac{6}{3}\mathbf{n}=2\mathbf{n}=\langle4, 2, -4\rangle\), the points are \(B_1=M+2\mathbf{n}=(6, 2, -5)\) and \(B_2=M-2\mathbf{n}=(-2, -2, 3)\). 3. Parallel tangent planes have the form \(2x+y-2z=d\). 4. Substituting \(B_1\) gives \(d=24\), and substituting \(B_2\) gives \(d=-12\). 5. Therefore, the tangent planes are \(T_1:2x+y-2z=24\) and \(T_2:2x+y-2z=-12\).

Points of tangency: \(B_1=(6, 2, -5)\) and \(B_2=(-2, -2, 3)\) Tangent planes: \(T_1:2x+y-2z=24\) and \(T_2:2x+y-2z=-12\)

- Complete the square to find the sphere's center and radius. - A plane perpendicular to a vector can use that vector as a normal vector. - Move one radius from the center in both normal directions. - A plane parallel to the y-axis has a normal vector with zero y-component.

1. Complete the square: \((x+2)^2+(y-1)^2+(z+3)^2=25\). Thus, the sphere has center \(M=(-2, 1, -3)\) and radius \(5\). 2. Because each plane is perpendicular to \(\mathbf{v}\), use \(\mathbf{v}\) as its normal vector. Its zero y-component also makes each plane parallel to the y-axis. 3. Since \(\|\mathbf{v}\|=5\), the tangent points are \(B_1=M+\mathbf{v}=(2, 1, -6)\) and \(B_2=M-\mathbf{v}=(-6, 1, 0)\). 4. The tangent planes have the form \(4x-3z=d\). Substitution gives \(d=26\) at \(B_1\) and \(d=-24\) at \(B_2\). 5. Therefore, the planes are \(T_1:4x-3z=26\) and \(T_2:4x-3z=-24\).

Points of tangency: \(B_1=(2, 1, -6)\) and \(B_2=(-6, 1, 0)\) Tangent planes: \(T_1:4x-3z=26\) and \(T_2:4x-3z=-24\)

- Tangency to a coordinate plane fixes the corresponding center coordinate at one radius from the plane. - Substitute the known point into the sphere equation. - Use the positive-coordinate condition to choose the valid solution.

1. Tangency to the yz-plane \(x=0\), together with positivity, gives \(m_1=9\). 2. Tangency to the xy-plane \(z=0\) gives \(m_3=9\). 3. Because \(P\) lies on the sphere, \((5-9)^2+(1-m_2)^2+(5-9)^2=81\). 4. Thus, \((1-m_2)^2=49\), so \(m_2=-6\) or \(m_2=8\). 5. The center must have positive coordinates, so \(m_2=8\). Therefore, \(M=(9,8,9)\).

\(M=(9,8,9)\)

- Tangency to all three coordinate planes makes all center coordinates equal to the radius. - Substitute the given point into the sphere equation. - Solve the resulting quadratic equation for the radius.

1. Because the sphere is tangent to all three coordinate planes in the first octant, its center has the form \(M(r,r,r)\), where \(r\) is the radius. 2. Since \(A\) lies on the sphere, \((1-r)^2+(2-r)^2+(1-r)^2=r^2\). 3. Expanding gives \(2r^2-8r+6=0\), or \(r^2-4r+3=0\). 4. Therefore, \(r=1\) or \(r=3\). 5. The possible centers are \(M_1=(1,1,1)\) and \(M_2=(3,3,3)\).

\(M_1=(1,1,1)\) and \(M_2=(3,3,3)\)

- Read the center and radius from the sphere equation. - Use the vector from the center to \(Q\) as the line direction. - Parameterize the entire line and impose the sphere equation; do not assume \(Q\) lies on the sphere.

1. The sphere has center \(M=(5,-2,1)\) and radius \(7\). 2. A direction vector from \(M\) toward \(Q\) is \(\langle4,6,12\rangle\), whose magnitude is \(14\). 3. Parameterize the line as \(\mathbf{r}(t)=\langle5,-2,1\rangle+t\langle4,6,12\rangle\). 4. Substituting in the sphere gives \(196t^2=49\), so \(t=\pm\frac12\). 5. The two intersection points are \((7,1,7)\) and \((3,-5,-5)\).

\((7,1,7)\) and \((3,-5,-5)\)

- The two centers and the tangency point are collinear. - For external tangency, the tangency point lies between the centers. - Move one new radius beyond the tangency point.

1. Sphere \(K\) has center \(M_K=(1,2,3)\) and radius \(5\). 2. The vector from \(M_K\) to the tangency point is \(\overrightarrow{M_KB}=\langle 0,4,3\rangle\), with magnitude \(5\). 3. For external tangency, the new center lies beyond \(B\) along the same ray and is \(10\) units from \(B\). 4. A length-\(10\) vector in that direction is \(2\langle 0,4,3\rangle=\langle 0,8,6\rangle\). 5. The new center is \(M_S=B+\langle 0,8,6\rangle=(1,14,12)\). 6. Therefore, the sphere is \((x-1)^2+(y-14)^2+(z-12)^2=100\).

\((x-1)^2+(y-14)^2+(z-12)^2=100\)

- Find the center and radius of \(K\). - The centers and tangency point are collinear. - Move one new radius from \(P\) in each direction along the center-to-point line.

1. Sphere \(K\) has center \(M_K=(1,-2,2)\) and radius \(12\). 2. The vector \(\overrightarrow{M_KP}=\langle 4,8,-8\rangle\) has magnitude \(12\). 3. A vector of length \(6\) in this direction is \(\frac{1}{2}\overrightarrow{M_KP}=\langle 2,4,-4\rangle\). 4. The two possible centers are \(M_1=P+\langle 2,4,-4\rangle=(7,10,-10)\) and \(M_2=P-\langle 2,4,-4\rangle=(3,2,-2)\). 5. Therefore, the sphere equations are \((x-7)^2+(y-10)^2+(z+10)^2=36\) and \((x-3)^2+(y-2)^2+(z+2)^2=36\).

\((x-7)^2+(y-10)^2+(z+10)^2=36\) and \((x-3)^2+(y-2)^2+(z+2)^2=36\)

- Verify that the stated point lies on both spheres. - Compare the distance between the centers with the sum of the radii. - The tangent plane is perpendicular to the radius at the point of tangency. - Use the line of centers as a normal direction for the plane.

1. The sphere centers are \(M_1=(2, 1, 4)\) and \(M_2=(10, 7, 4)\), and both radii are \(5\). 2. For \(B\), \(BM_1^2=4^2+3^2+0^2=25\) and \(BM_2^2=(-4)^2+(-3)^2+0^2=25\), so \(B\) lies on both spheres. 3. The center distance is \(M_1M_2=\sqrt{8^2+6^2}=10\), equal to the sum of the radii. Thus, the spheres are externally tangent at \(B\). 4. The common tangent plane is perpendicular to the line of centers. A normal vector is \(\overrightarrow{M_1M_2}=\langle8, 6, 0\rangle\), which is parallel to \(\langle4, 3, 0\rangle\). 5. The plane through \(B\) is \(4(x-6)+3(y-4)=0\), or \(4x+3y-36=0\).

The spheres are externally tangent at \(B=(6, 4, 4)\). Their common tangent plane is \(4x+3y-36=0\).

- The top vertices are vertical translations of the base vertices. - Use symmetry between the base and top planes. - The circumcenter of a right triangle is the midpoint of its hypotenuse.

1. Moving each base vertex vertically by \(6\) gives \(D(4,0,6)\), \(E(0,4,6)\), and \(F(0,0,6)\). 2. The sphere center lies halfway between the base and top planes, so its z-coordinate is \(3\). 3. In the xy-plane, the center is the circumcenter of right triangle \(ABC\), which is the midpoint of hypotenuse \(AB\). Thus, the center is \(M(2,2,3)\). 4. The squared radius is the squared distance from \(M\) to \(C\): \(r^2=2^2+2^2+3^2=17\). 5. The sphere equation is \((x-2)^2+(y-2)^2+(z-3)^2=17\).

a) \(D=(4,0,6)\), \(E=(0,4,6)\), \(F=(0,0,6)\) b) \((x-2)^2+(y-2)^2+(z-3)^2=17\)

- Use the diameter to find the radius. - Relate the center's height to the sphere's highest point. - Set \(z=0\) to find the ground-level cross section. - Use the circle circumference formula.

1. The sphere's radius is \(10\,\text{m}\). 2. Let the center be \((0,0,h)\). The highest point has z-coordinate \(h+10=16\), so \(h=6\). 3. The sphere equation is \(x^2+y^2+(z-6)^2=100\). 4. At ground level, \(z=0\). Substitution gives \(x^2+y^2+36=100\), so the ground circle has radius \(8\,\text{m}\). 5. Its circumference is \(2\pi(8)=16\pi\,\text{m}\).

a) \(x^2+y^2+(z-6)^2=100\) b) \(16\pi\,\text{m}\)

- Complete the square in all three variables. - Match the right side with \(r^2=49\). - Read the center from the completed-square form.

1. Complete the square: \((x-4)^2-16+(y+5)^2-25+(z+2)^2-4+c=0\). 2. Therefore, \((x-4)^2+(y+5)^2+(z+2)^2=45-c\). 3. The center is \(M=(4,-5,-2)\). 4. For radius \(7\), set \(45-c=49\). Thus, \(c=-4\).

\(c=-4\); center \(M=(4,-5,-2)\)

- First divide by the common coefficient of the squared terms. - Complete the square in each variable. - A real sphere requires a nonnegative squared radius.

1. Divide by \(3\): \(x^2+y^2+z^2+4x-6y+8z+31=0\). 2. Complete the square: \((x+2)^2-4+(y-3)^2-9+(z+4)^2-16+31=0\). 3. This becomes \((x+2)^2+(y-3)^2+(z+4)^2=-2\). 4. A sum of real squares cannot equal a negative number. Therefore, the equation has no real points and does not describe a sphere in \(\mathbb{R}^3\).

It does not describe a real sphere because the completed-square form is \((x+2)^2+(y-3)^2+(z+4)^2=-2\), which has no real solutions.

- Use the plane's normal vector to find the perpendicular distance from the center. - Compare that distance with the sphere's radius. - A plane that passes through a sphere without containing its center creates a circular cross-section. - Use the right triangle formed by the sphere radius, the plane distance, and the cross-section radius.

1. The plane's normal vector is \(\mathbf{n}=\langle2, -2, 1\rangle\), whose magnitude is \(3\). 2. The distance from \(M\) to the plane is \(d=\frac{|2(2)-2(-1)+4-1|}{3}=3\). 3. Since \(3<6\), the plane intersects the sphere in a circle. 4. The circle's radius is \(r_S=\sqrt{6^2-3^2}=\sqrt{27}=3\sqrt3\).

a) The plane intersects the sphere in a circle because \(3<6\). b) \(r_S=3\sqrt3\)

- Read the center and radius from the sphere equation. - A tangent plane is exactly one radius from the sphere's center. - Use the plane's normal vector to write the distance expression. - An absolute-value equation may produce two values.

1. The sphere has center \(M=(1, -2, 3)\) and radius \(5\). 2. The plane's normal vector is \(\mathbf{n}=\langle4, 0, -3\rangle\), whose magnitude is \(5\). 3. The distance from \(M\) to \(F_c\) is \(d=\frac{|4(1)-3(3)+c|}{5}=\frac{|c-5|}{5}\). 4. Tangency requires \(d=5\), so \(|c-5|=25\). 5. Therefore, \(c=30\) or \(c=-20\).

\(c=30\) or \(c=-20\)

- Read the sphere's center and radius from standard form. - Compare the center-to-plane distance with the radius. - Use the Pythagorean relationship for the cross-section radius. - Project the sphere's center onto the plane along the normal vector.

1. The sphere has center \(M=(1, 2, 3)\) and radius \(R=7\). 2. The plane's normal vector is \(\mathbf{n}=\langle2, -1, 2\rangle\), whose magnitude is \(3\). The distance from \(M\) to the plane is \(d=\frac{|2(1)-2+2(3)-12|}{3}=2\). 3. Since \(2<7\), the plane intersects the sphere in a circle. 4. The circle's radius is \(r=\sqrt{7^2-2^2}=\sqrt{45}=3\sqrt5\). 5. Its center is the perpendicular projection of \(M\) onto the plane. Write \(M_s=M+t\mathbf{n}\). Substitution gives \(6+9t=12\), so \(t=\frac23\). 6. Therefore, \(M_s=\left(\frac73, \frac43, \frac{13}{3}\right)\).

The plane intersects the sphere in a circle. \(r=3\sqrt5\) \(M_s=\left(\frac73, \frac43, \frac{13}{3}\right)\)

- Complete the square in all three variables. - A sphere with positive radius requires its squared radius to be positive. - In part c), use the given radius to write an equation for the unknown parameter.

1. Complete the square to obtain \((x-2k)^2+(y+1)^2+(z-3)^2=4k^2-3\). 2. Therefore, \(M_k=(2k,-1,3)\) and \(r^2=4k^2-3\). 3. A sphere with positive radius requires \(4k^2-3>0\), so \(\lvert k\rvert>\frac{\sqrt{3}}{2}\). 4. For radius \(5\), set \(4k^2-3=25\). Then \(k^2=7\), so \(k=\pm\sqrt{7}\).

a) \(M_k=(2k,-1,3)\), \(r^2=4k^2-3\) b) \(\lvert k\rvert>\frac{\sqrt{3}}{2}\) c) \(k=\pm\sqrt{7}\)

- Complete the square in x and z. - Express the center coordinates using the parameter. - Tangency to the xy-plane means the absolute z-coordinate of the center equals the radius.

1. Complete the square: \((x-2a)^2+y^2+(z-a)^2=16\). 2. Thus, every member is a sphere of radius \(4\), with center \(M_a=(2a,0,a)\). 3. The centers lie on the line \((x,y,z)=t\langle 2,0,1\rangle\). 4. A sphere is tangent to the xy-plane when the center's distance from \(z=0\) equals the radius. Therefore, \(\lvert a\rvert=4\), so \(a=4\) or \(a=-4\).

a) \((x-2a)^2+y^2+(z-a)^2=16\). The squared radius is always positive, so every real \(a\) gives a sphere with radius \(4\). b) \((x,y,z)=t\langle2,0,1\rangle\) c) \(a=4\) or \(a=-4\)

- Read the center and radius from standard form. - Points on the x-axis have y- and z-coordinates equal to \(0\). - Substitute the known coordinates of \(S\).

1. The center is \(M=(-1,4,0)\), and the radius is \(5\). 2. Points on the x-axis have \(y=z=0\). Substitution gives \((x+1)^2+16=25\), so \((x+1)^2=9\). 3. Therefore, \(x=2\) or \(x=-4\), giving \((2,0,0)\) and \((-4,0,0)\). 4. Substitute \(S\): \(3^2+4^2+z^2=25\). Thus, \(z^2=0\), so \(z=0\).

a) Center \((-1,4,0)\); radius \(5\) b) \((2,0,0)\) and \((-4,0,0)\) c) \(z=0\)

- Use the center-to-surface point distance for the radius. - Substitute \(Q\) into the sphere equation. - Compare the center's distance from the xy-plane with the radius.

1. The squared radius is \((5-3)^2+(-1+2)^2+(2-4)^2=9\), so \(r=3\). 2. The sphere equation is \((x-3)^2+(y+2)^2+(z-4)^2=9\). 3. For \(Q\), the left side is \((1-3)^2+(-4+2)^2+(5-4)^2=9\), so \(Q\) lies on the sphere. 4. The center is \(4\) units from the xy-plane, while the radius is only \(3\). Therefore, the sphere has no points in the xy-plane.

a) \((x-3)^2+(y+2)^2+(z-4)^2=9\) b) \(Q\) lies on the sphere. c) The center-to-plane distance is \(4>3\), so there is no intersection.

- Complete the square separately for the \(x\)-, \(y\)-, and \(z\)-terms. - Compare the result with \((x-a)^2+(y-b)^2+(z-c)^2=r^2\). - Use the distance formula in three dimensions.

1. Group the terms by variable and complete each square: \((x^2-4x+4)+(y^2+6y+9)+(z^2-2z+1)=2+4+9+1\). 2. This gives \((x-2)^2+(y+3)^2+(z-1)^2=16\), which is the standard equation of a sphere. 3. Therefore, the center is \(M=(2,-3,1)\), and the radius is \(r=4\). 4. The requested distance equals \(MO=\sqrt{(2-0)^2+(-3-0)^2+(1-0)^2}=\sqrt{14}\).

a) \((x-2)^2+(y+3)^2+(z-1)^2=16\) b) \(M=(2,-3,1)\) and \(r=4\) c) \(\sqrt{14}\)

- Use the three-dimensional distance formula for each point. - A sphere with center \((h, k, \ell)\) and radius \(r\) has equation \((x-h)^2+(y-k)^2+(z-\ell)^2=r^2\). - Substitute the coordinates of \(P\) into the sphere equation. - Use the inequality in the problem to choose between the two possible values.

1. Compute the squared distances from \(M\): \(MA^2=4^2+0^2+(-3)^2=25\), \(MB^2=0^2+5^2+0^2=25\), and \(MC^2=3^2+4^2+0^2=25\). 2. Thus, all three points are \(5\) units from \(M\), so the sphere equation is \((x-2)^2+(y+1)^2+(z-5)^2=25\). 3. Substitute \(P=(2, y, 8)\): \((y+1)^2+3^2=25\), so \((y+1)^2=16\). 4. Therefore, \(y=3\) or \(y=-5\). The condition \(y>-1\) gives \(y=3\).

a) Each distance is \(5\). b) \((x-2)^2+(y+1)^2+(z-5)^2=25\) c) \(y=3\)

- Compare each point's squared distance from the origin with \(100\). - Points in the xy-plane have \(z=0\). - Points on the z-axis have \(x=y=0\). - Recall the standard equation of a circle centered at the origin.

1. Compare \(x^2+y^2+z^2\) with \(100\). For \(P\), the value is \(36+0+64=100\), so \(P\) is on the sphere. For \(Q\), it is \(49+49+1=99\), so \(Q\) is inside. 2. For \(R\), the value is \(100+1+0=101\), so \(R\) is outside. For \(S\), it is \(50+25+25=100\), so \(S\) is on the sphere. 3. In the xy-plane, \(z=0\). The intersection therefore satisfies \(x^2+y^2=100\), a circle centered at the origin with radius \(10\). 4. On the z-axis, \(x=y=0\). Then \(z^2=100\), so \(z=\pm10\).

a) \(P\) and \(S\) are on the sphere; \(Q\) is inside; \(R\) is outside. b) A circle with equation \(x^2+y^2=100\) c) \((0, 0, 10)\) and \((0, 0, -10)\)

- A sphere tangent to the three coordinate planes has equal positive center coordinates. - Write the fourth face using its x-, y-, and z-intercepts. - Use the face's normal vector to compute the perpendicular distance from the center. - Set that distance equal to the sphere's radius.

1. Because the sphere is tangent to the three coordinate planes and lies in the first octant, its center is \(M=(r, r, r)\). 2. The face through \(X\), \(Y\), and \(Z\) has intercept equation \(\frac{x}{3}+\frac{y}{3}+\frac{z}{6}=1\), or \(2x+2y+z=6\). A normal vector is \(\mathbf{n}=\langle2, 2, 1\rangle\), with magnitude \(3\). 3. The distance from \(M\) to this face must also equal \(r\). Since \(M\) lies below the face, \(r=\frac{6-(2r+2r+r)}{3}\). 4. Therefore, \(3r=6-5r\), so \(8r=6\) and \(r=\frac34\).

\(r=\frac34\)

- Complete the square to find the sphere's center and radius. - Use the plane's normal vector to find the center-to-plane distance. - The sphere radius, cross-section radius, and center-to-plane distance form a right triangle. - The cross-section's center is the perpendicular projection of the sphere's center onto the plane.

1. Complete the square: \((x-1)^2+(y+2)^2+(z-3)^2=49\). Thus, the sphere has center \(M=(1, -2, 3)\) and radius \(7\). 2. The plane has normal vector \(\mathbf{n}=\langle3, 0, -4\rangle\), whose magnitude is \(5\). The distance from \(M\) to the plane is \(d=\frac{|3(1)-4(3)-12|}{5}=\frac{21}{5}\). 3. Since \(\frac{21}{5}<7\), the plane intersects the sphere in a circle. Its radius is \(\rho=\sqrt{7^2-\left(\frac{21}{5}\right)^2}=\frac{28}{5}=5.6\). 4. The center \(Z\) is the perpendicular projection of \(M\) onto the plane. Write \(Z=M+\lambda\mathbf{n}=(1+3\lambda, -2, 3-4\lambda)\). 5. Substitution into \(3x-4z=12\) gives \(25\lambda=21\), so \(\lambda=\frac{21}{25}\). Therefore, \(Z=\left(\frac{88}{25}, -2, -\frac{9}{25}\right)=(3.52, -2, -0.36)\).

The plane intersects the sphere in a circle. \(\rho=\frac{28}{5}=5.6\) \(Z=\left(\frac{88}{25}, -2, -\frac{9}{25}\right)=(3.52, -2, -0.36)\)

- Compare the center-to-plane distance with the sphere's radius. - Use the right triangle formed by the sphere radius, plane distance, and cross-section radius. - The intersection circle's center is the projection of the sphere's center onto the plane. - A line in the plane misses the sphere when it lies outside the intersection circle.

1. The sphere has center \(M=(-1, 4, 2)\) and radius \(10\). The plane's normal vector is \(\mathbf{n}=\langle3, 0, -4\rangle\), whose magnitude is \(5\). 2. The distance from \(M\) to the plane is \(d=\frac{|3(-1)-4(2)+31|}{5}=4\). Since \(4<10\), the intersection is a circle. 3. Its radius is \(r_S=\sqrt{10^2-4^2}=\sqrt{84}=2\sqrt{21}\). 4. The circle's center is the perpendicular projection of \(M\) onto \(E\). Write \(M_S=M+t\mathbf{n}\). Substitution gives \(20+25t=0\), so \(t=-\frac45\) and \(M_S=\left(-\frac{17}{5}, 4, \frac{26}{5}\right)\). 5. The vectors \(\langle0, 1, 0\rangle\) and \(\langle4, 0, 3\rangle\) both lie parallel to \(E\) and are perpendicular to each other. Let \(P=M_S+2\langle4, 0, 3\rangle=\left(\frac{23}{5}, 4, \frac{56}{5}\right)\). 6. One possible line is \(g:\mathbf{x}=P+s\langle0, 1, 0\rangle\). Its distance from \(M_S\) is \(10>2\sqrt{21}\), so it lies outside the intersection circle and does not meet the sphere.

a) \(d(M,E)=4<10\), so the intersection is a circle. b) \(r_S=2\sqrt{21}\) c) One possible line is \(g:\mathbf{x}=\left\langle\frac{23}{5}, 4, \frac{56}{5}\right\rangle+s\langle0, 1, 0\rangle\). Its distance from the intersection circle's center is \(10>r_S\).

- Complete the square to write the sphere in standard form. - Compare the center-to-plane distance with the radius. - The nearest point on the sphere lies along the perpendicular from the center toward the plane. - Subtract the radius from the center-to-plane distance.

1. Complete the square: \(x^2+(y-3)^2+(z+4)^2=25\). Thus, \(M=(0, 3, -4)\) and \(r=5\). 2. The plane's normal vector is \(\mathbf{n}=\langle2, 1, -2\rangle\), whose magnitude is \(3\). The distance from \(M\) to the plane is \(d=\frac{|2(0)+3-2(-4)+5|}{3}=\frac{16}{3}\). 3. Since \(\frac{16}{3}>5\), the plane does not intersect the sphere. 4. The point on the sphere closest to the plane lies from \(M\) in the direction opposite \(\mathbf{n}\), because the plane expression is positive at \(M\). Therefore, \(P=M-5\frac{\mathbf{n}}{\|\mathbf{n}\|}\). 5. This gives \(P=(0, 3, -4)-\frac53\langle2, 1, -2\rangle=\left(-\frac{10}{3}, \frac43, -\frac23\right)\). 6. The minimum distance from the sphere to the plane is \(d-r=\frac{16}{3}-5=\frac13\).

a) \(M=(0, 3, -4)\) and \(r=5\) b) The plane does not intersect the sphere because \(\frac{16}{3}>5\). c) \(P=\left(-\frac{10}{3}, \frac43, -\frac23\right)\), and the minimum distance is \(\frac13\).

- Find both centers and exact radii. - Compare the center distance with the sum and difference of the radii. - The tangent point lies on the line through the two centers. - The common tangent plane is perpendicular to that line.

1. The centers are \(M_1=(1, 2, 3)\) and \(M_2=(3, 4, 5)\). The radii are \(r_1=4\sqrt3\) and \(r_2=2\sqrt3\). 2. The center distance is \(M_1M_2=\sqrt{2^2+2^2+2^2}=2\sqrt3\). Since this equals \(|r_1-r_2|\), the spheres are internally tangent. 3. The point of tangency lies on the line of centers. Because \(r_1=2M_1M_2\), move twice the vector \(\overrightarrow{M_1M_2}=\langle2, 2, 2\rangle\) from \(M_1\): \(B=(1, 2, 3)+2\langle2, 2, 2\rangle=(5, 6, 7)\). 4. The tangent plane is perpendicular to the line of centers, so a normal vector is \(\langle1, 1, 1\rangle\). 5. The plane through \(B\) is \((x-5)+(y-6)+(z-7)=0\), or \(x+y+z-18=0\).

a) The spheres are internally tangent because \(M_1M_2=|r_1-r_2|=2\sqrt3\). b) \(B=(5, 6, 7)\) c) \(x+y+z-18=0\)

- Substitute the four vertices into a general sphere equation. - Complete the square to identify the center and radius. - For one sphere to lie inside another, compare the center distance plus the smaller radius with the larger radius.

1. Let \(K_1\) have equation \(x^2+y^2+z^2+ax+by+cz+d=0\). 2. Substituting \(O\) gives \(d=0\). Substituting \(P\), \(Q\), and \(R\) gives \(a=b=c=-8\). 3. Thus, \(x^2+y^2+z^2-8x-8y-8z=0\). Completing the square gives \((x-4)^2+(y-4)^2+(z-4)^2=48\). 4. Therefore, \(M_1=(4,4,4)\) and \(r_1=4\sqrt{3}\). 5. Sphere \(K_2\) has center \(M_2=(2,2,2)\) and radius \(2\). The center distance is \(M_1M_2=2\sqrt{3}\). 6. Since \(2\sqrt{3}+2<4\sqrt{3}\), every point of \(K_2\) is strictly inside \(K_1\).

a) \(K_1: (x-4)^2+(y-4)^2+(z-4)^2=48\); center \((4,4,4)\); radius \(4\sqrt{3}\) b) \(M_1M_2+r_2=2\sqrt{3}+2<4\sqrt{3}=r_1\), so \(K_2\) lies completely inside \(K_1\).

- Use a point on the base rim and the highest point of the dome. - Both points are one sphere radius from the center. - Compare the instrument's squared distance from the center with \(R^2\).

1. A point on the base rim is \(A(24,0,0)\), and the dome's top is \(H(0,0,18)\). 2. Let the sphere center be \(M(0,0,c)\). Equal distances from \(M\) to \(A\) and \(H\) give \(24^2+c^2=(18-c)^2\). 3. Solving gives \(c=-7\). Then \(R^2=24^2+7^2=625\), so \(R=25\,\text{m}\). 4. The sphere equation is \(x^2+y^2+(z+7)^2=625\). 5. For \(P\), the squared distance to the center is \(15^2+10^2+(25+7)^2=1349\). 6. Since \(1349>625\), the instrument lies outside the sphere.

a) Radius \(25\,\text{m}\); equation \(x^2+y^2+(z+7)^2=625\) b) The instrument lies outside the sphere.

- The radius equals the center-to-plane distance. - The point of tangency is the perpendicular projection of the center onto the plane. - Use the plane's normal vector as the direction of the perpendicular line. - Under reflection, the plane is the perpendicular bisector of the segment joining the original and reflected centers.

1. The radius equals the distance from \(M\) to the tangent plane: \(r=\frac{|1-10|}{\sqrt{1^2+(-2)^2+2^2}}=\frac93=3\). 2. The point of tangency is the perpendicular projection of \(M\) onto \(E\). Write \(B=(1,0,0)+t\langle1,-2,2\rangle\). 3. Substitution into the plane equation gives \(1+9t-10=0\), so \(t=1\). Therefore, \(B=(2,-2,2)\). 4. The reflected center lies the same distance beyond \(B\), so \(M'=M+2\overrightarrow{MB}\). 5. Since \(\overrightarrow{MB}=\langle1,-2,2\rangle\), \(M'=(1,0,0)+2\langle1,-2,2\rangle=(3,-4,4)\).

a) \(r=3\) b) \(B=(2, -2, 2)\) c) \(M'=(3, -4, 4)\)

- The tangent radius equals the shortest distance from the center to the line. - Use a dot product to locate the point on the line closest to the center. - If the radius is greater than that distance, the line intersects the sphere twice. - Substitute the line into the sphere equation to find the intersection parameters.

1. A general point on \(g\) is \(P(t)=(2+t, 3+2t, 1+2t)\). Then \(\overrightarrow{MP(t)}=\langle t-3, 2t, 2t-3\rangle\). 2. At the closest point, this vector is perpendicular to the direction vector \(\langle1, 2, 2\rangle\). Thus, \((t-3)+2(2t)+2(2t-3)=9t-9=0\), so \(t=1\). 3. The closest point is \((3, 5, 3)\), whose distance from \(M\) is \(\sqrt{(-2)^2+2^2+(-1)^2}=3\). Therefore, the tangent sphere has radius \(r=3\). 4. For \(r=5\), substitute the line into \((x-5)^2+(y-3)^2+(z-4)^2=25\). This gives \(9t^2-18t-7=0\). 5. Solving gives \(t=-\frac13\) or \(t=\frac73\). The corresponding points are \(\left(\frac53, \frac73, \frac13\right)\) and \(\left(\frac{13}{3}, \frac{23}{3}, \frac{17}{3}\right)\). 6. Since there are two intersection points, \(g\) is a secant when \(r=5\).

a) \(r=3\) b) For \(r=5\), \(g\) is a secant. The intersection points are \(\left(\frac53, \frac73, \frac13\right)\) and \(\left(\frac{13}{3}, \frac{23}{3}, \frac{17}{3}\right)\).