Free math worksheets

- Focus on the radius that ends at the intended tangency point. - What relationship must a tangent have with that radius? - The required line must pass through \(T\).

1. Draw the line through \(T\) perpendicular to radius \(\overline{OT}\). 2. A line perpendicular to a radius at its endpoint on the circle is tangent to the circle. 3. Therefore, the tangent makes a \(90^\circ\) angle with \(\overline{OT}\).

Construct the line through \(T\) perpendicular to \(\overline{OT}\). It is tangent at \(T\), and the angle with \(\overline{OT}\) is \(90^\circ\).

- Compare each point's distance from the center with the radius. - Consider separately what happens for a point outside, on, and inside a circle. - A tangent touches the circle at exactly one point.

1. Point \(P\) is outside the circle because \(7>5\), so two tangent lines can be constructed through \(P\). 2. Point \(Q\) lies on the circle because \(5=5\), so exactly one tangent line can be constructed through \(Q\). 3. Point \(R\) is inside the circle because \(3<5\), so no tangent line to the circle can pass through \(R\).

Through \(P\): \(2\) tangents Through \(Q\): \(1\) tangent Through \(R\): \(0\) tangents

- What geometric condition must hold at a true point of tangency? - Ask whether Lena's equal-length condition forces that angle condition. - What familiar circle theorem guarantees a right angle when \(OP\) is used as a diameter?

1. Lena's auxiliary circle guarantees that each intersection point \(T\) satisfies \(PT=PO\), but that equality does not force \(OT\perp PT\). Therefore, it does not guarantee that \(PT\) is tangent to the original circle. 2. Construct the midpoint \(M\) of \(\overline{OP}\). 3. Draw the auxiliary circle centered at \(M\) with diameter \(\overline{OP}\). 4. If this circle intersects the original circle at \(T_1\) and \(T_2\), then \(\angle OT_1P\) and \(\angle OT_2P\) are right angles because they subtend diameter \(OP\). 5. Hence \(OT_i\perp PT_i\), so \(PT_1\) and \(PT_2\) are the required tangents.

Centering the auxiliary circle at \(P\) does not force a right angle between a radius and the proposed tangent. Instead, construct the circle with diameter \(OP\). Its intersections \(T_1\) and \(T_2\) with the given circle satisfy \(\angle OT_iP=90^\circ\), so \(PT_1\) and \(PT_2\) are tangent.

- A radius is perpendicular to a tangent at the point of tangency. - What auxiliary circle would force a right angle at each desired tangency point? - Describe the construction in terms of the midpoint of \(MQ\) and the intersections of two circles. - After locating a tangency point, identify a right triangle containing \(M\), \(Q\), and that point.

1. The diagram shows circle \(k\) with center \(M\), radius \(3\,\text{cm}\), and \(MQ=8\,\text{cm}\). 2. Construct midpoint \(K\) of \(\overline{MQ}\), and draw the auxiliary circle centered at \(K\) with diameter \(\overline{MQ}\). 3. Let the auxiliary circle intersect \(k\) at \(T_1\) and \(T_2\). Draw lines \(QT_1\) and \(QT_2\). Each is tangent because an angle subtending diameter \(MQ\) is a right angle, so \(MT_i\perp QT_i\). 4. In right triangle \(MT_1Q\), \(MT_1^2+T_1Q^2=MQ^2\). 5. Substitute: \(3^2+T_1Q^2=8^2\), so \(T_1Q^2=55\). 6. Therefore, \(T_1Q=T_2Q=\sqrt{55}\,\text{cm}\approx7.42\,\text{cm}\).

Construct the midpoint \(K\) of \(\overline{MQ}\), draw the circle with diameter \(MQ\), and let its intersections with \(k\) be \(T_1\) and \(T_2\). The required tangent lines are \(QT_1\) and \(QT_2\). Each tangent segment has length \(\sqrt{55}\,\text{cm}\approx7.42\,\text{cm}\).

- A tangent is perpendicular to the radius at the point of tangency. - What auxiliary circle would make an angle at each desired tangency point a right angle? - Describe the construction using the midpoint of \(MP\) and the intersections of two circles. - For part b, classify \(Q\) as outside, on, or inside the original circle.

1. Construct the midpoint \(Z\) of \(\overline{MP}\). 2. Draw the circle with diameter \(\overline{MP}\). Let its intersections with circle \(k\) be \(T_1\) and \(T_2\). 3. Draw lines \(PT_1\) and \(PT_2\). Because each angle subtending diameter \(\overline{MP}\) is a right angle, \(MT_1\perp PT_1\) and \(MT_2\perp PT_2\). A line perpendicular to a radius at its endpoint is tangent to the circle. 4. If \(MQ>r\), then \(Q\) is outside the circle and there are two tangents. 5. If \(MQ=r\), then \(Q\) is on the circle and there is one tangent, perpendicular to \(\overline{MQ}\) at \(Q\). 6. If \(MQ<r\), then \(Q\) is inside the circle and there are no tangent lines through \(Q\).

a) Construct the circle with diameter \(MP\). If its intersections with \(k\) are \(T_1\) and \(T_2\), then \(PT_1\) and \(PT_2\) are the tangents because \(\angle MT_iP=90^\circ\). b) \(MQ>r\): two tangents; \(MQ=r\): one tangent; \(MQ<r\): no tangents.

- How is a circle's diameter related to its radius? - In part b), use only the fact that \(T_1\) lies on the auxiliary circle with diameter \(\overline{MA}\) before deciding anything about tangency. - After establishing the right angle, what tangent criterion applies to radius \(\overline{MT_1}\)? - Compare the possible locations of \(A\) with the interior, boundary, and exterior of the original circle.

1. Since \(\overline{MA}\) is the diameter of the auxiliary circle, \(MA=2\cdot5.5=11\,\text{cm}\). 2. Point \(T_1\) lies on the auxiliary circle with diameter \(\overline{MA}\), so \(\angle MT_1A=90^\circ\) by the inscribed-angle theorem for a diameter. 3. Therefore, \(\overline{AT_1}\perp\overline{MT_1}\). Since \(MT_1\) is a radius of the original circle, a line perpendicular to that radius at \(T_1\) is tangent to the original circle. Thus \(\overline{AT_1}\) is tangent. 4. A tangent from \(A\) is impossible when \(A\) lies inside the original circle. If \(A\) lies on the circle, exactly one tangent exists.

a) \(MA=11\,\text{cm}\) b) \(\angle MT_1A=90^\circ\) because it is an inscribed angle intercepting diameter \(\overline{MA}\) of the auxiliary circle. Hence \(AT_1\perp MT_1\), so \(\overline{AT_1}\) is tangent to the original circle at \(T_1\). c) \(A\) must lie inside the original circle.

- The auxiliary circle's diameter is \(\overline{MP}\). - Its center is the midpoint of that diameter. - Imagine the tangent lines from a point very far from the circle.

1. Since \(P\) lies on the outer circle, \(MP = 6\,\text{cm}\). 2. The auxiliary circle has \(\overline{MP}\) as its diameter, so its radius is \(3\,\text{cm}\). 3. Point \(K\) is the midpoint of \(\overline{MP}\), so \(MK = 3\,\text{cm}\). 4. As \(P\) moves farther from \(M\), the tangent lines become closer to parallel. The tangent points move away from the ray \(MP\) toward the endpoints of the diameter perpendicular to \(\overline{MP}\). 5. Therefore, the tangent points move farther apart, and the minor arc between them increases toward a semicircle.

a) The auxiliary-circle radius is \(3\,\text{cm}\). b) \(MK = 3\,\text{cm}\). c) The tangent points move farther apart toward the endpoints of the diameter perpendicular to \(\overline{MP}\); the minor arc between them increases.

- Determine whether \(g\) is horizontal or vertical. - Recall the angle between a tangent and the radius to its point of tangency. - Lines perpendicular to the same line are parallel. - Move one radius from the center in both directions along \(g\).

1. Line \(g\) is the vertical line \(x = 5\). Tangents perpendicular to \(g\) must be horizontal. 2. A tangent is perpendicular to the radius at the point of tangency. Therefore, the radii to \(T_1\) and \(T_2\) must lie on \(g\). 3. Move \(3\) units up and down from \(M(5, 5)\): \(5 + 3 = 8\) and \(5 - 3 = 2\). 4. The points of tangency are \(T_1(5, 8)\) and \(T_2(5, 2)\). The tangent lines are \(y = 8\) and \(y = 2\).

1. The tangents are the horizontal lines \(y = 8\) and \(y = 2\). 2. \(T_1(5, 8)\) and \(T_2(5, 2)\)

- Begin with the segment joining the center to the point on the circle. - Consider the distance from the center to any other point on the perpendicular line. - Use the right triangle formed by the center, the tangency point, and that other point.

1. Line \(t\) passes through \(T\) and is perpendicular to radius \(\overline{OT}\). 2. For any other point \(Q\) on \(t\), triangle \(OTQ\) is right at \(T\), so \(OQ^2=OT^2+TQ^2\). 3. Since \(Q\ne T\), \(TQ^2>0\), and therefore \(OQ>OT\). Thus every other point on \(t\) lies outside the circle. 4. Hence \(t\) meets the circle only at \(T\) and is the tangent at \(T\).

For any other point \(Q\) on \(t\), the right triangle \(OTQ\) gives \(OQ^2=OT^2+TQ^2>OT^2\), so \(Q\) lies outside the circle. Thus \(T\) is the only point of \(t\) on the circle, and \(t\) is tangent at \(T\).

- Use the equal-tangent-segments theorem at external point \(P\). - Identify another point that is equidistant from \(A\) and \(B\). - Two distinct points on the perpendicular bisector determine that line.

1. The app draws the line through \(A\) perpendicular to \(OA\) and the line through \(B\) perpendicular to \(OB\). These are the tangents at \(A\) and \(B\), and they meet at \(P\). 2. Tangent segments from the same external point are congruent, so \(PA=PB\). Therefore, \(P\) lies on the perpendicular bisector of \(\overline{AB}\). 3. Radii \(OA\) and \(OB\) are congruent, so \(O\) is also equidistant from \(A\) and \(B\). Therefore, \(O\) lies on the same perpendicular bisector. 4. Since the perpendicular bisector is the unique line through the two distinct points \(O\) and \(P\), line \(OP\) is the perpendicular bisector of \(\overline{AB}\).

Because \(PA=PB\) and \(OA=OB\), both \(P\) and \(O\) lie on the perpendicular bisector of \(AB\). Hence \(OP\) is that perpendicular bisector.

- Look for a construction that forces a right angle at a point on the given circle. - Use \(\overline{OP}\) as the diameter of an auxiliary circle. - Relate a right angle between a radius and a segment to tangency.

1. Construct the midpoint \(M\) of \(\overline{OP}\). 2. Draw the circle centered at \(M\) with radius \(MO\). This circle has \(\overline{OP}\) as a diameter. 3. Let the two intersections of this auxiliary circle with the original circle be \(T_1\) and \(T_2\). 4. Because \(\overline{OP}\) is a diameter of the auxiliary circle, \(\angle OT_1P\) and \(\angle OT_2P\) are right angles. 5. Thus \(OT_1\perp PT_1\) and \(OT_2\perp PT_2\). A line perpendicular to a radius at its endpoint on the circle is tangent, so \(\overline{PT_1}\) and \(\overline{PT_2}\) are the two tangent segments.

Construct the circle with diameter \(\overline{OP}\). Its intersections \(T_1\) and \(T_2\) with the given circle determine the tangent segments \(\overline{PT_1}\) and \(\overline{PT_2}\), because each forms a right angle with the corresponding radius.

- Tangency at a specified point determines a line on which the center must lie. - Tangency to both intersecting lines requires equal perpendicular distances from the center. - Combine the two center loci and count their intersections.

1. Construct the line \(n\) through \(T\) perpendicular to \(\ell\). The center of any circle tangent to \(\ell\) at \(T\) must lie on \(n\). 2. Construct the internal and external angle-bisector lines of \(\ell\) and \(m\). A point on either bisector is equidistant from the two lines. 3. Let \(C_1\) and \(C_2\) be the intersections of \(n\) with the two angle-bisector lines. 4. Draw the circle centered at \(C_i\) with radius \(C_iT\), for \(i=1,2\). Because \(C_iT\perp\ell\), each circle is tangent to \(\ell\) at \(T\). 5. Since each \(C_i\) lies on an angle bisector, its perpendicular distance to \(m\) equals its distance \(C_iT\) to \(\ell\). Thus each circle is also tangent to \(m\). 6. Any valid center must lie on both \(n\) and one of the two angle-bisector lines. These intersections are exactly \(C_1\) and \(C_2\), so there are exactly two circles.

Intersect the perpendicular to \(\ell\) at \(T\) with the internal and external angle-bisector lines of \(\ell\) and \(m\). The two intersection points are the centers; use each center's distance to \(T\) as its radius.

- Replace the two unequal radii with their difference. - First find a direction tangent from one center to the reduced circle. - How can a parallel offset restore the smaller circle's radius while keeping the tangent direction?

1. Draw the auxiliary circle centered at \(O_1\) with radius \(5-2=3\,\text{cm}\). 2. Construct the two tangents from \(O_2\) to the auxiliary circle. Let one tangent touch it at \(T\). 3. Radius \(O_1T\) is perpendicular to tangent \(O_2T\). 4. On ray \(O_1T\), mark point \(V\) so that \(O_1V=5\,\text{cm}\). Through \(V\), construct line \(n\parallel O_2T\). 5. Drop the perpendicular from \(O_2\) to \(n\), meeting \(n\) at \(U\). Since the distance between the parallel lines \(O_2T\) and \(n\) is \(O_1V-O_1T=5-3=2\,\text{cm}\), \(O_2U=2\,\text{cm}\). 6. Line \(n\) is perpendicular to both radii \(O_1V\) and \(O_2U\), so it is tangent to both original circles. 7. Repeating the construction with the other auxiliary tangent gives the second common external tangent.

Shrinking the larger radius by the smaller radius reduces the problem to tangents from \(O_2\) to a circle of radius \(3\,\text{cm}\). Parallel offsets of those two tangent directions produce the two common external tangents.