- 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

- Sort the points into outside, on, and inside. - What is special about the tangent at a given point of the circle? - Could a line through an interior point touch only once?

1. Point \(P\) is outside the circle, so two tangent lines can be drawn through it. 2. Point \(Q\) lies on the circle, so exactly one tangent can be drawn at \(Q\). 3. Point \(R\) is inside the circle, so every line through it meets the circle again; no tangent is possible.

\(P\): two tangents, because the point is outside the circle. \(Q\): one tangent, because the point lies on the circle. \(R\): no tangents, because every line through an interior point crosses the circle twice.

- 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.

- Read the radius direction from the marked angle. - What turn from a radius gives a tangent direction? - Compare both proposed directions without relying on a pre-drawn tangent.

1. Radius \(OT\) points \(35^\circ\) above the positive horizontal direction. 2. A tangent at \(T\) must make a \(90^\circ\) angle with \(OT\), so one valid line direction is \(35^\circ+90^\circ=125^\circ\). 3. Line \(g\) has this direction; line \(h\) does not. Thus \(g\) is the tangent.

Line \(g\), because the difference between the line's direction and the radius direction is \(90^\circ\).

- How does the radius meet a tangent line at their contact point? - Which circle fact makes an inscribed angle a right angle? - Think about the endpoints of the diameter opposite that right angle.

1. A tangent is perpendicular to the radius at the touching point, so \(\angle OTP=90^\circ\). 2. A point \(T\) on a circle with diameter \(OP\) has \(\angle OTP=90^\circ\) by the inscribed-angle theorem. 3. The auxiliary circle has center at the midpoint of \(OP\) and passes through both \(O\) and \(P\).

\(\angle OTP=90^\circ\). Use \(OP\) as the auxiliary circle's diameter, centered at the midpoint of \(OP\).

- 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 \(T_2(5, 2)\)

- Pick a second point anywhere on the perpendicular line. - Compare that point's distance from the center with the radius. - Why does an extra side of a right triangle increase its hypotenuse?

1. Choose any other point \(Q\) on line \(t\). 2. Triangle \(OTQ\) is right at \(T\), so \(OQ^2=OT^2+TQ^2\). 3. As \(Q\ne T\), the length \(TQ\) is positive. Hence \(OQ>OT\), and \(Q\) lies outside the circle. 4. Therefore, \(T\) is the only point shared by line \(t\) and the circle, making \(t\) tangent.

For every other point \(Q\) of \(t\), \(OQ^2=OT^2+TQ^2>OT^2\). Thus \(Q\) is outside the circle and \(t\) meets it only at \(T\).

- Which two lengths are equal because they are tangent segments from \(P\)? - What other pair is equal because the segments are radii? - How can two equidistant points determine a perpendicular bisector?

1. The tangent segments from the same external point are congruent, so \(PA=PB\). 2. Thus \(P\) is on the perpendicular bisector of \(AB\). 3. Equal radii give \(OA=OB\), so \(O\) is also on that perpendicular bisector. 4. The line through the distinct points \(O\) and \(P\) is therefore the perpendicular bisector of \(AB\).

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

- What does tangency tell you about the distance from a center to the tangent line? - Think about which side of the tangent each center must lie on. - When a line through \(O_2\) shifts, how do its distances from the two centers change?

1. A tangent from \(O_2\) to a circle of radius \(10\,\text{cm}\) around \(O_1\) is \(10\,\text{cm}\) from \(O_1\). Shifting it \(3\,\text{cm}\) toward \(O_1\) puts the two original centers on opposite sides of the new line. That constructs a common internal tangent, not the requested external tangent. 2. For both centers to remain on the same side, the auxiliary radius must be \(7-3=4\,\text{cm}\). 3. Construct the two tangents from \(O_2\) to the circle centered at \(O_1\) with radius \(4\,\text{cm}\). For each tangent, move a parallel copy \(3\,\text{cm}\) away from \(O_1\). The new line is \(7\,\text{cm}\) from \(O_1\) and \(3\,\text{cm}\) from \(O_2\), with both centers on the same side, so it touches both original circles.

a) A radius of \(10\,\text{cm}\) gives common internal tangents: the original centers end up on opposite sides of each tangent. b) Use radius \(4\,\text{cm}\). Draw tangents from \(O_2\) to that auxiliary circle and move parallel copies \(3\,\text{cm}\) away from \(O_1\).

- Where must a center lie when one contact point is known? - What locus describes points equidistant from two intersecting lines? - Find common points of the two conditions rather than trying different circles.

1. Every possible center must lie on the line \(n\) through \(T\) perpendicular to \(\ell\). 2. A circle tangent to both intersecting lines has its center on one of the two angle-bisector lines of \(\ell\) and \(m\). 3. Intersect \(n\) with each of these two bisectors, obtaining centers \(C_1\) and \(C_2\). 4. A circle with center \(C_i\) and radius \(C_iT\) touches \(\ell\) at \(T\). Because \(C_i\) is equidistant from the two lines, it also touches \(m\). 5. The perpendicular through \(T\) intersects each of the two bisectors exactly once, giving two centers and no others.

Intersect the perpendicular to \(\ell\) at \(T\) with both angle-bisector lines of \(\ell\) and \(m\). Each intersection is a center, with radius equal to its distance from \(T\). There are exactly two.

- Think about how far a common tangent lies from each center. - Could a parallel line through one center be easier to construct first? - How would you move that line back to make it touch both circles?

1. Draw an auxiliary circle centered at \(O_1\), with radius \(5-2=3\,\text{cm}\). 2. Construct the two tangents from \(O_2\) to this auxiliary circle. Let one touch it at \(T\). 3. On ray \(O_1T\), choose point \(V\) so that \(O_1V=5\,\text{cm}\), then draw through \(V\) the line \(n\) parallel to \(O_2T\). 4. The distance between these parallel lines is \(5-3=2\,\text{cm}\), so line \(n\) is also tangent to the smaller circle centered at \(O_2\). 5. Since \(n\perp O_1V\), it is tangent to the larger circle at \(V\). Repeat with the second auxiliary tangent to get the other common external tangent.

First construct tangents from \(O_2\) to a circle centered at \(O_1\) with radius \(5-2=3\,\text{cm}\). Offset each tangent outward by \(2\,\text{cm}\) using a parallel line through a point on the larger circle. The two offset lines are the common external tangents.