- A point lies on a curve when its coordinates satisfy the curve's equation. - Test the two points separately in \(x^2+y^2=1\). - Compare each resulting value with the right side of the equation.

1. Substitute \(P(0.8, 0.6)\): \((0.8)^2+(0.6)^2=0.64+0.36=1\). Therefore, \(P\) lies on the circle. 2. Substitute \(Q(1, 1)\): \(1^2+1^2=2\ne1\). Therefore, \(Q\) does not lie on the circle.

Only \(P(0.8, 0.6)\) lies on the unit circle.

- Match each squared binomial to \(x-h\) or \(y-k\). - Pay attention to the sign inside the y-binomial. - The number on the right is \(r^2\), not \(r\).

1. Compare the equation with standard form \((x-h)^2+(y-k)^2=r^2\). 2. Since \(x-4=x-h\), \(h=4\). Since \(y+3=y-(-3)\), \(k=-3\). 3. Since \(r^2=49\), the radius is \(r=7\).

Center: \((4,-3)\); radius: \(7\)

- What is the standard form for a circle centered at \((h,k)\)? - Take care with the sign of a negative center coordinate.

1. Use \((x-h)^2+(y-k)^2=r^2\). 2. Substitute \(h=-2\), \(k=3\), and \(r=4\) to get \((x+2)^2+(y-3)^2=16\).

\((x+2)^2+(y-3)^2=16\)

- Find the center directly from the squared terms. - Compare the distance from each graph's center to its rightmost point. - Check both center and radius before deciding.

1. The equation has center \((2,-1)\) and radius \(\sqrt{9}=3\). 2. Graph a) has that center and radius. Graph b) moves the center to \((-2,1)\); graph c) uses radius \(2\).

a), because its center is \((2,-1)\) and its radius is \(3\).

- Compare each point's squared distance from the center with the radius squared. - What does equality mean? What do smaller and larger values mean?

1. The circle has squared radius \(25\). 2. For \(A\), \(3^2+4^2=25\), so \(A\) is on the circle. 3. For \(B\), \(1^2+2^2=5<25\), so \(B\) is inside. 4. For \(C\), \(6^2+0^2=36>25\), so \(C\) is outside.

\(A\): on. \(B\): inside. \(C\): outside.

- How does the center-radius form simplify when the center is the origin? - What y-coordinate must an x-axis intersection have? - For part c), substitute the specified x-coordinate into the equation you wrote and keep both square roots.

1. The circle has equation \(x^2+y^2=25\). 2. On the x-axis, \(y=0\), so \(x^2=25\). Thus, the intercepts are \((5,0)\) and \((-5,0)\). 3. For \(x=2\), substitute into the equation: \(2^2+y^2=25\), so \(y^2=21\). 4. Therefore, \(y=\sqrt{21}\) or \(y=-\sqrt{21}\).

a) \(x^2+y^2=25\) b) \((5,0)\) and \((-5,0)\) c) \(y=\sqrt{21}\) and \(y=-\sqrt{21}\)

- Use the center and radius to write the circle in center-radius form. - At an intersection, the point must have the power line's fixed y-coordinate and also satisfy the circle equation. - Keep both square roots when solving for \(x\).

1. The circle centered at the origin with radius \(10\) has equation \(x^2+y^2=100\). 2. At an intersection with the power line, \(y=5\). Substitute: \(x^2+5^2=100\). 3. Thus \(x^2=75\), so \(x=5\sqrt{3}\) or \(x=-5\sqrt{3}\). 4. The intersection points are \((5\sqrt{3},5)\) and \((-5\sqrt{3},5)\).

a) \(x^2+y^2=100\) b) \((5\sqrt{3},5)\) and \((-5\sqrt{3},5)\)

- Check whether two sides of the triangle are perpendicular. - Where is the circumcenter of a right triangle? - Once the center is known, how do you find the radius and standard equation?

1. Segments \(AB\) and \(AC\) are horizontal and vertical, so \(\angle BAC=90^\circ\). 2. The center of the circle through the three triangle vertices is the midpoint of the hypotenuse \(BC\): \(\left(\frac{5+1}{2},\frac{1+5}{2}\right)=(3,3)\). 3. The radius is the distance from \((3,3)\) to \((1,1)\), or \(\sqrt{2^2+2^2}=2\sqrt{2}\). 4. The equation is \((x-3)^2+(y-3)^2=8\).

Center: \((3,3)\). Radius: \(2\sqrt{2}\). Equation: \((x-3)^2+(y-3)^2=8\). Since \(ABC\) is right-angled at \(A\), its circumcenter is the midpoint of hypotenuse \(BC\).

- Use the wheel's center and radius in the standard equation of a circle. - Substitute the specified height for \(y\) before solving for the horizontal coordinate. - The two square roots represent positions on opposite sides of the vertical centerline.

1. The circle has center \((0,20)\) and radius \(18\), so its equation is \(x^2+(y-20)^2=324\). 2. At a height of \(29\,\text{ft}\), substitute \(y=29\): \(x^2+9^2=324\). 3. Thus, \(x^2=243\), so \(x=\pm9\sqrt{3}\). 4. Therefore, the rider is about \(15.6\,\text{ft}\) to either side of the centerline.

a) \(x^2+(y-20)^2=324\) b) \(x=\pm9\sqrt{3}\,\text{ft}\), approximately \(15.6\,\text{ft}\) to either side of the centerline

- Use the leftmost and rightmost points of the circle to locate the horizontal midpoint. - Check the highest and lowest points to locate the vertical midpoint and radius. - Once you know \((h,k)\) and \(r\), use the standard center-radius form of a circle.

1. The horizontal extremes are at \(x=-5\) and \(x=1\), so their midpoint has x-coordinate \(-2\). The vertical extremes are at \(y=-2\) and \(y=4\), so their midpoint has y-coordinate \(1\). 2. Therefore, the center is \((-2,1)\). 3. The distance from the center to any extreme point is \(3\), so the radius is \(3\). 4. Substitute \(h=-2\), \(k=1\), and \(r=3\) into \((x-h)^2+(y-k)^2=r^2\): \((x+2)^2+(y-1)^2=9\).

a) Center: \((-2,1)\); radius: \(3\) b) \((x+2)^2+(y-1)^2=9\)

- Rewrite each binomial mentally in the form \(x-h\) or \(y-k\). - Ask what value of \(h\) makes \(x-h=x+5\). - In standard form, what does the constant on the right represent: \(r\) or \(r^2\)?

1. Standard form is \((x-h)^2+(y-k)^2=r^2\). 2. The expression \(x+5\) is \(x-(-5)\), so the center's x-coordinate is \(-5\), not \(5\). The expression \(y-2\) gives y-coordinate \(2\), not \(-2\). 3. The right side is \(r^2=16\), so \(r=4\), not \(16\). 4. The correct center is \((-5,2)\) and the radius is \(4\).

The signs inside the binomials are opposite the center coordinates, and the right side is the radius squared. The correct center is \((-5,2)\) and the radius is \(4\).

- Identify the center and radius from the equation. - Which coordinate changes at the farthest left and right points? - Which coordinate changes at the top and bottom points?

1. The center is \((-1,2)\), and the radius is \(3\). 2. Move three units horizontally from the center: left \((-4,2)\), right \((2,2)\). 3. Move three units vertically: top \((-1,5)\), bottom \((-1,-1)\). 4. These four points lie on the circle along the horizontal and vertical diameters and locate its outer extent.

Center: \((-1,2)\). Left: \((-4,2)\); right: \((2,2)\); top: \((-1,5)\); bottom: \((-1,-1)\). These points mark the circle's left, right, upper, and lower edges, so they guide a sketch with radius \(3\).

- Group the terms that contain \(x\) and those that contain \(y\). - Complete each square separately and balance both sides. - Read the center only after obtaining standard form.

1. Group the variables: \((x^2-4x)+(y^2+6y)=3\). 2. Add \(4\) and \(9\) to both sides to complete the squares: \((x-2)^2+(y+3)^2=16\). 3. The center is \((2,-3)\), and the radius is \(4\).

\((x-2)^2+(y+3)^2=16\). Center: \((2,-3)\); radius: \(4\).

- What distance must every point on the circle have from the center? - Compare the x- and y-coordinates separately. - Use the distance as the radius in standard form.

1. The horizontal and vertical differences from the center to \(P\) are \(4\) and \(3\). 2. The radius is \(\sqrt{4^2+3^2}=5\). 3. Thus the equation is \((x+1)^2+(y-2)^2=25\).

Radius: \(5\). Equation: \((x+1)^2+(y-2)^2=25\).

- Find the centers by comparing the extreme points of both circles. - Which direction is negative along the x-axis? Which is positive along the y-axis? - Translate the new center coordinates into squared terms.

1. The gray circle is centered at the origin. 2. The blue circle is centered at \((-3,2)\), so it moved \(3\) units left and \(2\) units up. 3. Its standard equation is \((x+3)^2+(y-2)^2=4\).

The blue circle moved \(3\) units left and \(2\) units up. Its equation is \((x+3)^2+(y-2)^2=4\).

- Read the center and radius from the equation first. - A vertical line is a fixed horizontal distance from the center. - Compare that distance with the radius for each case.

1. The circle's center is \((2,-1)\), and its radius is \(4\). 2. A vertical line \(x=a\) is \(|a-2|\) units from the center. 3. The distances for a), b), and c) are \(4\), \(2\), and \(5\), respectively. 4. Distance equal to the radius means tangent, less means secant, and greater means no intersection.

a) Tangent. b) Secant. c) Does not meet the circle.

- Apply the zero-product property to the second equation. - Substitute each resulting fixed value into the other equation. - List each distinct ordered pair only once.

1. By the zero-product property, \(x-3=0\) or \(y-4=0\). Thus, consider the cases \(x=3\) and \(y=4\). 2. If \(x=3\), then \(9+y^2=25\), so \(y=4\) or \(y=-4\). This gives \((3,4)\) and \((3,-4)\). 3. If \(y=4\), then \(x^2+16=25\), so \(x=3\) or \(x=-3\). This gives \((3,4)\) and \((-3,4)\). 4. Remove the repeated ordered pair \((3,4)\).

\(\{(3,4),(3,-4),(-3,4)\}\)

- Solve the linear equation for one variable. - Substitute into the quadratic equation and expand. - Factor the resulting quadratic, then find each matching coordinate.

1. Solve the linear equation for \(x\): \(x=y+5\). 2. Substitute into the quadratic equation: \((y+5)^2+y^2=13\). 3. Expand and simplify: \(2y^2+10y+12=0\). Divide by \(2\): \(y^2+5y+6=0\). 4. Factor: \((y+2)(y+3)=0\). Thus, \(y=-2\) or \(y=-3\). 5. Using \(x=y+5\), the corresponding values are \(x=3\) and \(x=2\).

\(\{(3,-2),(2,-3)\}\)

- A point of intersection must satisfy both the circle and line equations. - One equation is already solved for a variable. What happens when you substitute it into the other? - Check each candidate point in both original equations.

1. Substitute the line equation into the circle equation: \((2y-4)^2+y^2=13\). 2. Expand and simplify: \(5y^2-16y+3=0\). 3. Factor: \((5y-1)(y-3)=0\), so \(y=\frac{1}{5}\) or \(y=3\). 4. If \(y=\frac{1}{5}\), then \(x=2\cdot\frac{1}{5}-4=-\frac{18}{5}\). If \(y=3\), then \(x=2\). 5. Both ordered pairs satisfy the original circle and line equations.

\(\left\{\left(-\frac{18}{5},\frac{1}{5}\right),(2,3)\right\}\)

- Solve the linear equation for one variable. - Substitute into the second equation. - Factor the resulting quadratic and find each corresponding coordinate.

1. Solve the linear equation for \(y\): \(y=1-x\). 2. Substitute into the second equation: \(x^2+(1-x)^2=13\). 3. Expand and simplify: \(2x^2-2x-12=0\). Divide by \(2\): \(x^2-x-6=0\). 4. Factor: \((x-3)(x+2)=0\). Thus, \(x=3\) or \(x=-2\). 5. Using \(y=1-x\), the corresponding values are \(y=-2\) and \(y=3\).

\(\{(3,-2),(-2,3)\}\)

- Every point on the unit circle is a distance \(1\) from the origin. - Use the Pythagorean relationship between \(x\) and \(y\). - Use the quadrant condition when choosing a square root. - A point lies on the circle only if its coordinates satisfy \(x^2+y^2=1\).

1. Use \(x^2+y^2=1\): \(x^2+0.44^2=1\), so \(x^2=0.8064\). Because \(P\) is in Quadrant I, \(x=\sqrt{0.8064}\approx0.898\). 2. Substitute \(x=\frac{3}{5}\): \(\frac{9}{25}+y^2=1\), so \(y^2=\frac{16}{25}\). In Quadrant I, \(y=\frac{4}{5}\). 3. For \(Q\), \(0.6^2+0.7^2=0.36+0.49=0.85\ne1\). Therefore, \(Q\) is not on the unit circle.

1. \(x\approx0.898\) 2. \(y=\frac{4}{5}\) 3. No, because \(0.6^2+0.7^2=0.85\ne1\).

- Read the center and radius from the squared expressions. - How far is the horizontal chord from the center? - What right triangle can you form with a radius and half the chord?

1. The center is \(C(3,-1)\) and the radius is \(4\). 2. The line \(y=1\) is \(2\) units above the center. Its perpendicular from \(C\) meets the chord at its midpoint. 3. A right triangle formed by the radius, the distance \(2\), and half the chord gives \(\left(\frac{AB}{2}\right)^2=4^2-2^2=12\). 4. Therefore \(AB=2\sqrt{12}=4\sqrt{3}\) units. The chord midpoint lies directly above \(C\) on \(y=1\), at \((3,1)\).

a) Center \((3,-1)\); radius \(4\). b) \(AB=4\sqrt{3}\) units. c) Midpoint \((3,1)\), directly above the center because the perpendicular from a circle's center bisects a chord.

- Compare each equation with the standard form of a circle. - What happens to the squared terms if you compare the two equations? - Where must the midpoint of a diameter lie?

1. The first circle has center \((0,0)\) and radius \(8\). The second has center \((4,4)\) and radius \(\sqrt{32}=4\sqrt{2}\). 2. Expand the second equation: \(x^2+y^2-8x-8y+32=32\), so \(x^2+y^2=8x+8y\). 3. Using \(x^2+y^2=64\), get \(x+y=8\). Then \(y=8-x\), and substitution gives \(x^2+(8-x)^2=64\). 4. Simplifying gives \(2x(x-8)=0\), so the intersection points are \((0,8)\) and \((8,0)\). 5. Their midpoint is \((4,4)\), the center of the second circle. Thus they are endpoints of one of its diameters, but not a diameter of the first circle, whose center is \((0,0)\).

a) First circle: center \((0,0)\), radius \(8\). Second circle: center \((4,4)\), radius \(4\sqrt{2}\). b) \((0,8)\) and \((8,0)\). c) They form a diameter of the second circle only, since their midpoint is \((4,4)\).

- Complete the squares to reveal the center and the radius expression. - Compare the radius with the center's perpendicular distance to the x-axis. - Locate the tangency point along that perpendicular direction.

1. Complete the squares: \((x-3)^2+(y+2)^2=13-k\). 2. The center is \((3, -2)\), whose distance from the x-axis is \(2\). 3. Tangency to the x-axis requires the radius to equal that distance, so \(r=2\) and \(r^2=4\). 4. Therefore \(13-k=4\), giving \(k=9\). 5. The tangency point lies vertically above the center at \((3, 0)\).

\(k=9\) Center: \((3, -2)\) Radius: \(2\) Tangency point: \((3, 0)\)

- Draw horizontal and vertical legs from the center to a general point \(P(x,y)\). - Express each leg length using coordinate differences. - How is the distance from the center to every point on the circle related to \(r\)? - For part b, first determine \(r^2\) from the center and the given point.

1. From the center \((h,k)\) to \(P(x,y)\), the horizontal and vertical changes are \(x-h\) and \(y-k\). 2. These changes are the legs of a right triangle whose hypotenuse is the radius \(r\). By the Pythagorean theorem, \((x-h)^2+(y-k)^2=r^2\). 3. For the given circle, the squared radius is the squared distance from \((-3,2)\) to \((1,5)\): \(r^2=(1+3)^2+(5-2)^2=4^2+3^2=25\). 4. Substitute \(h=-3\), \(k=2\), and \(r^2=25\): \((x+3)^2+(y-2)^2=25\).

a) \((x-h)^2+(y-k)^2=r^2\) b) \((x+3)^2+(y-2)^2=25\)

- Group the x-terms and y-terms before completing either square. - Whatever constants you add to create perfect-square trinomials must also be added to the other side of the equation. - After finding standard form, substitute the line's fixed y-value and count the real intersection points.

1. Move the constant term: \(x^2-6x+y^2+8y=11\). 2. Complete each square by adding \(9\) and \(16\) to both sides: \((x-3)^2+(y+4)^2=11+9+16=36\). 3. Therefore, the center is \((3,-4)\) and the radius is \(6\). 4. Substitute \(y=2\): \((x-3)^2+(2+4)^2=36\), so \((x-3)^2=0\) and \(x=3\). 5. The line has exactly one intersection point, \((3,2)\), so \(y=2\) is tangent to the circle.

a) \((x-3)^2+(y+4)^2=36\) b) Center: \((3,-4)\); radius: \(6\) c) \(y=2\) is tangent at \((3,2)\).

- Solve the linear equation for \(x\). - Substitute and expand the squared binomial. - Factor the resulting quadratic in \(y\), then find the matching \(x\)-values.

1. Solve the linear equation for \(x\): \(x=y+c\). 2. Substitute into the first equation: \((y+c)^2+y^2=25c^2\). 3. Expand and simplify: \(2y^2+2cy-24c^2=0\). Divide by \(2\): \(y^2+cy-12c^2=0\). 4. Factor: \((y-3c)(y+4c)=0\). Thus, \(y=3c\) or \(y=-4c\). 5. Using \(x=y+c\), the ordered pairs are \((4c,3c)\) and \((-3c,-4c)\). 6. When \(c=0\), these expressions represent the same ordered pair, \((0,0)\).

a) \(y^2+cy-12c^2=0\) b) For \(c\ne0\), \(\{(4c,3c),(-3c,-4c)\}\). For \(c=0\), \(\{(0,0)\}\).

- Write the circle equation and identify the parabola's vertex. - Use the condition that the vertex lies on the circle. - Rewrite one equation so that \(x^2\) can be substituted into the other.

1. The circle has equation \(x^2+y^2=100\), and the parabola has vertex \((0,c)\). 2. For the vertex to lie on the circle, \(c^2=100\). Since \(c<0\), \(c=-10\). 3. The system is \(x^2+y^2=100\) and \(y=\frac{1}{4}x^2-10\). Rewrite the second equation as \(x^2=4y+40\). 4. Substitute into the circle equation: \(4y+40+y^2=100\), so \(y^2+4y-60=0\). 5. Factor: \((y+10)(y-6)=0\). Thus, \(y=-10\) or \(y=6\). 6. When \(y=-10\), \(x^2=0\), giving \((0,-10)\). When \(y=6\), \(x^2=64\), giving \((8,6)\) and \((-8,6)\).

a) \(c=-10\) b) \(\{(0,-10),(8,6),(-8,6)\}\)

- What must the center's two coordinates be if the circle touches both axes? - Write the equation using just the radius as an unknown. - Check whether each possible radius puts the center in Quadrant I.

1. If the radius is \(r>0\), tangency to both axes means the center is \((r,r)\). 2. Point \(P\) lies on the circle, so \((4-r)^2+(2-r)^2=r^2\). 3. Simplifying gives \(r^2-12r+20=0\), or \((r-2)(r-10)=0\). 4. Both \(r=2\) and \(r=10\) are positive. Each gives a center in Quadrant I and a circle through \(P\). 5. Their equations are \((x-2)^2+(y-2)^2=4\) and \((x-10)^2+(y-10)^2=100\).

\((x-2)^2+(y-2)^2=4\) \((x-10)^2+(y-10)^2=100\) Both circles have positive radii and centers \((r,r)\), and both pass through \(P(4,2)\).