- Recall the theorem about tangent segments from one external point. - Compare the two segments that begin at \(P\) and end at the tangency points. - Use the length marked in the diagram only after identifying the relationship.

1. Tangent segments drawn from the same external point to a circle are congruent. 2. The diagram gives \(PA=8.4\,\text{cm}\). Therefore, \(PA=PB\), so \(PB=8.4\,\text{cm}\).

\(PB=8.4\,\text{cm}\)

- Focus on the radius that ends at the point of tangency. - What angle does a tangent form with that radius? - The vertex of \(\angle OTP\) is the tangency point \(T\).

1. A tangent to a circle is perpendicular to the radius drawn to the point of tangency. 2. Therefore, \(\overline{OT}\perp\overline{PT}\), so \(m\angle OTP=90^\circ\).

\(90^\circ\)

- Identify the chord and the center of the circle. - Which circle theorem uses the marked right angle?

1. A perpendicular from the center of a circle to a chord cuts the chord in half. 2. Since \(OM\perp AB\), point \(M\) bisects \(AB\), so \(AM=MB\).

\(AM=MB\). The perpendicular from the center to a chord bisects the chord.

- Compare how many times each line meets the circle. - Look at the radius that reaches the contact point. - How is a tangent oriented relative to that radius?

1. Line \(g\) meets the circle at point \(T\) and is perpendicular to radius \(OT\). 2. A line perpendicular to a radius at its endpoint on the circle is tangent there. 3. Line \(h\) crosses the circle in two places, so it is a secant.

Line \(g\) is tangent at \(T\) because \(g\perp OT\). Line \(h\) is a secant.

- A radius to a point of tangency is perpendicular to the tangent line. - Identify the equation of the line through \(P\) and \(Q\). - For a vertical diameter, only the y-coordinate changes from the center.

1. The radius is the horizontal distance from \(M(3, 3)\) to the y-axis, so \(r = |3 - 0| = 3\). The point of tangency is \(B(0, 3)\). 2. The line through \(P(0, 0)\) and \(Q(7, 0)\) is the x-axis, \(y = 0\). 3. The perpendicular distance from \(M(3, 3)\) to the x-axis is \(3\). Since this distance equals the radius, \(w\) is tangent to the circle. 4. A vertical diameter lies on \(x = 3\). Moving \(3\) units above and below the center gives \(K(3, 6)\) and \(L(3, 0)\).

a) \(r = 3\) and \(B(0, 3)\). b) \(w\) is tangent because its distance from \(M\) is \(3\), equal to the radius. c) One choice is \(K(3, 6)\) and \(L(3, 0)\).

- Convert the lengths to the same units. - Compare each shortest distance with the radius. - How many intersection points result when a line lies inside, at, or beyond that radius?

1. Convert the distances to centimeters: \(3.5\,\text{cm}\), \(4.5\,\text{cm}\), and \(5\,\text{cm}\). 2. Since \(3.5<4.5\), line \(g_1\) crosses the circle at two points and is a secant. 3. Since \(4.5=4.5\), line \(g_2\) meets the circle once and is a tangent. 4. Since \(5>4.5\), line \(g_3\) has no point on the circle.

\(g_1\): secant; its distance is less than the radius. \(g_2\): tangent; its distance equals the radius. \(g_3\): no intersection; its distance exceeds the radius.

- Picture two parallel lines that just touch opposite sides of a circle. - Consider the radii from the center to the two points of tangency. - Relate the resulting segment to the circle’s diameter.

1. A tangent line is perpendicular to the radius drawn to its point of tangency. 2. Because the two tangents are parallel, the two perpendicular radii lie on the same line through the center and form a diameter. 3. Therefore, the distance between the tangent lines equals the diameter: \(2r = 2 \cdot 4.5\,\text{cm} = 9\,\text{cm}\).

The tangent lines are \(9\,\text{cm}\) apart. Their points of tangency are endpoints of a diameter, so the distance is \(2 \cdot 4.5\,\text{cm} = 9\,\text{cm}\).

- Which segment is a radius to the point of tangency? - What theorem determines the angle between that radius and the tangent? - After establishing the right angle, use the two labeled lengths to relate them to \(MP\).

1. A radius drawn to a point of tangency is perpendicular to the tangent, so \(\triangle MTP\) is a right triangle. 2. The diagram shows legs \(MT=5\,\text{cm}\) and \(TP=12\,\text{cm}\). Apply the Pythagorean theorem: \(MP^2=5^2+12^2=169\). 3. Therefore, \(MP=13\,\text{cm}\).

a) Radius \(MT\) is perpendicular to the tangent at \(T\), so \(\angle MTP=90^\circ\). b) \(MP=13\,\text{cm}\)

- Which chord is shared by the marked inscribed angle and the tangent-chord angle? - Recall how an angle formed by a tangent and a chord is related to an inscribed angle intercepting the same chord. - Use the numerical angle shown in the diagram after identifying that relationship.

1. By the Tangent-Chord Theorem, the angle formed by a tangent and a chord equals an inscribed angle that intercepts the same chord. 2. The diagram gives \(m\angle BAC=42^\circ\). Therefore, \(m\angle CBE=m\angle BAC=42^\circ\).

\(42^\circ\)

- Which segment is a radius to the point where the tangent touches the circle? - What relationship must that radius have to the tangent? - After justifying the angle at \(B\), use the numerical angle marked in the diagram and the triangle angle sum.

1. A radius to a point of tangency is perpendicular to the tangent, so \(m\angle OBA=90^\circ\). 2. The diagram gives \(m\angle OAB=32^\circ\). The angle measures of triangle \(OBA\) sum to \(180^\circ\). 3. Therefore, \(m\angle AOB=180^\circ-90^\circ-32^\circ=58^\circ\).

a) Radius \(OB\) is perpendicular to the tangent at \(B\), so \(\angle OBA=90^\circ\). b) \(m\angle AOB=58^\circ\)

- A radius is perpendicular to a tangent at the point of tangency. - Consider the angle sum of quadrilateral \(AMBQ\). - Two of the quadrilateral's angles are right angles; use the remaining marked angle from the diagram.

1. Radius \(\overline{MA}\) is perpendicular to tangent segment \(\overline{QA}\), and radius \(\overline{MB}\) is perpendicular to tangent segment \(\overline{QB}\). Thus, \(\angle MAQ=90^\circ\) and \(\angle MBQ=90^\circ\). 2. The diagram gives \(m\angle AMB=120^\circ\). The interior angles of quadrilateral \(AMBQ\) sum to \(360^\circ\). 3. Therefore, \(120^\circ+90^\circ+\alpha+90^\circ=360^\circ\). 4. Solving gives \(\alpha=60^\circ\).

\(\alpha=60^\circ\)

- What angle does a tangent make with the radius to its point of contact? - How are \(OA\) and \(OB\) positioned when \(AB\) is a diameter? - What can you say about two distinct lines perpendicular to one line?

1. A tangent to a circle is perpendicular to the radius at its point of contact. 2. Thus the tangent at \(A\) is perpendicular to \(OA\), and the tangent at \(B\) is perpendicular to \(OB\). 3. Both radii lie on the diameter line \(AB\). Distinct lines perpendicular to the same line are parallel.

The tangent–radius perpendicularity theorem says a tangent is perpendicular to the radius at the point of contact. Both radii lie on the diameter \(AB\), so both tangents are perpendicular to \(AB\) and therefore parallel.

- Which two distances are equal within each circle? - Where does a point equidistant from the endpoints of a segment lie? - How many lines pass through two distinct points?

1. Both \(O_1A\) and \(O_1B\) are radii of the first circle, so \(O_1A=O_1B\). 2. Hence \(O_1\) lies on the perpendicular bisector of \(\overline{AB}\). 3. Likewise, \(O_2A=O_2B\), so \(O_2\) lies on the same perpendicular bisector. 4. The two distinct centers determine that line, so \(O_1O_2\) is the perpendicular bisector of \(\overline{AB}\).

Each center is equidistant from \(A\) and \(B\), so both lie on the chord's perpendicular bisector. Their connecting line is that perpendicular bisector.

- The center must be equidistant from the two sides of the angle. - Use the perpendicular through \(A\) to identify one center-to-side distance. - Recognize the special right triangle formed with the angle bisector.

1. The angle bisector makes \(45^\circ\) with each ray. Triangle \(OAC\) is a \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle because \(AC\perp OX\). 2. Since \(OA=4\,\text{cm}\), the equal legs give \(AC=4\,\text{cm}\). 3. Point \(C\) lies on the angle bisector, so its perpendicular distances to the two rays are equal. 4. One of those distances is \(AC=4\,\text{cm}\), so both distances are \(4\,\text{cm}\). 5. Therefore, the circle centered at \(C\) with radius \(4\,\text{cm}\) is tangent to both rays and lies inside the angle.

Triangle \(OAC\) is a \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle, so \(AC=OA=4\,\text{cm}\). Since \(C\) lies on the angle bisector, it is \(4\,\text{cm}\) from both rays. Thus the circle is tangent to both rays inside the angle.

- What angle does a radius make with a tangent at its contact point? - Which side do the two triangles share? - What can triangle congruence tell you about the two angles at \(P\)?

1. A radius to a point of tangency is perpendicular to the tangent, so \(OT\perp PT\) and \(OU\perp PU\). 2. The right triangles \(PTO\) and \(PUO\) have shared hypotenuse \(PO\) and equal legs \(OT=OU\), so they are congruent by HL. 3. Therefore, \(\angle TPO=\angle OPU\), and ray \(\overrightarrow{PO}\) bisects \(\angle TPU\).

\(\triangle PTO\cong\triangle PUO\) by HL because the radii are perpendicular to the tangents. Thus \(\angle TPO=\angle OPU\).

- Compare the two triangles formed by the center and the congruent chords. - Identify the equal radii and the shared segment. - Use the resulting triangle congruence to compare the two angles at \(A\).

1. In triangles \(AOB\) and \(AOC\), \(AB=AC\) by the given chord congruence. 2. Also, \(OB=OC\) because they are radii, and \(AO\) is shared. 3. Therefore, the triangles are congruent by SSS. 4. Corresponding angles \(\angle BAO\) and \(\angle OAC\) are congruent. 5. Hence ray \(\overrightarrow{AO}\) bisects \(\angle BAC\).

Triangles \(AOB\) and \(AOC\) are congruent by SSS, so \(\angle BAO=\angle OAC\). Therefore, \(\overrightarrow{AO}\) is the angle bisector.

- Pair the two pieces of each complete chord. - What product relationship holds when two chords intersect inside a circle? - Substitute the three labeled lengths only after writing that relationship.

1. For two chords intersecting inside a circle, the products of the two segment lengths are equal: \(AP\cdot PB=CP\cdot PD\). 2. From the diagram, \(AP=3\), \(PB=8\), and \(CP=4\). Let \(PD=x\). 3. Then \(3\cdot8=4x\), so \(24=4x\). 4. Therefore, \(PD=6\).

\(PD=6\)

- Locate the arc intercepted by \(\angle TVU\). - What does the tangent-chord theorem say about an angle at \(T\)? - Which portion of the circle lies inside each angle?

1. Central angle \(TOU\) measures \(120^\circ\), so minor arc \(TU\) measures \(120^\circ\). 2. Inscribed angle \(TVU\) intercepts the same minor arc, so it measures \(60^\circ\). 3. The tangent-chord theorem says the angle between the tangent and chord on that side also measures half the intercepted arc, or \(60^\circ\).

Both angles measure \(60^\circ\).

- Draw the radii to the two points of tangency and connect \(M\) to \(P\). - What equal lengths do you know from radii and from tangent segments drawn from one external point? - Compare the two triangles on either side of \(\overline{MP}\); what does triangle congruence tell you about the angle at \(P\)? - After that, identify a right triangle that contains the radius and \(MP\).

1. Let the tangent points be \(S\) and \(T\). Then \(MS=MT\) because both are radii, \(PS=PT\) because tangent segments from the same external point are congruent, and \(MP\) is common to triangles \(MPS\) and \(MPT\). 2. By SSS, \(\triangle MPS\cong\triangle MPT\). Therefore, \(MP\) bisects the \(60^\circ\) angle between the tangents, so each resulting angle at \(P\) is \(30^\circ\). 3. A radius to a point of tangency is perpendicular to the tangent. In right triangle \(MPT\), \(\sin(30^\circ)=\frac{MT}{MP}=\frac{2}{MP}\). 4. Since \(\sin(30^\circ)=\frac{1}{2}\), \(MP=4\,\text{cm}\).

\(MP=4\,\text{cm}\)

- A radius and tangent are perpendicular at the point of tangency. - The two acute angles in a right triangle are complementary. - Compare a fixed opposite leg with an increasing adjacent leg.

1. A tangent is perpendicular to the radius at the point of tangency, so \(MB \perp BA\). Therefore, \(\angle MBA = 90^\circ\), and the right-angle vertex is \(B\). 2. The acute angles of a right triangle are complementary, so \(\angle MAB = 90^\circ - 58^\circ = 32^\circ\). 3. As \(A\) moves farther from \(B\), leg \(AB\) increases while leg \(MB\) remains fixed. 4. From vertex \(A\), the ratio of the opposite leg to the adjacent leg decreases. Therefore, \(\angle MAB\) decreases and approaches \(0^\circ\).

a) The triangle is right at \(B\) because the tangent is perpendicular to radius \(\overline{MB}\). b) \(\angle MAB = 32^\circ\). c) \(\angle MAB\) decreases toward \(0^\circ\) as \(A\) moves farther from \(B\).

- Use the top view to identify the observer and the two pillar centers on one line. - Replace each diameter with its radius before comparing the similar sightline triangles. - Express the observer's distance to the farther center in terms of the distance to the nearer center.

1. The pillar radii are \(r_1=0.60\,\text{m}\) and \(r_2=0.90\,\text{m}\). 2. Let \(x\) be the distance from the observer to the center of the smaller pillar. The distance to the center of the larger pillar is \(x+6\). 3. When the pillars have the same apparent width, the right triangles formed by a sightline and a radius to each point of tangency are similar. Therefore, \(\frac{r_1}{x}=\frac{r_2}{x+6}\). 4. Substitute: \(\frac{0.60}{x}=\frac{0.90}{x+6}\). 5. Cross-multiply: \(0.60(x+6)=0.90x\), so \(0.60x+3.60=0.90x\). 6. Thus, \(3.60=0.30x\), and \(x=12\,\text{m}\).

The observer must stand \(12\,\text{m}\) from the center of the smaller pillar.

- How does the perpendicular from the center split a chord? - Find the center-to-chord distance with the radius and half-chord. - In which diagram does the triangle's altitude pass through the center?

1. A perpendicular from the center to the chord bisects it into two pieces of \(4\,\text{cm}\). 2. The center, chord midpoint, and chord endpoint form a right triangle. The distance from the center to the chord is \(\sqrt{5^2-4^2}=3\,\text{cm}\). 3. In a), the vertex is on the far side of the center, so the triangle's altitude is \(5+3=8\,\text{cm}\). A congruent side is \(\sqrt{8^2+4^2}=4\sqrt{5}\,\text{cm}\). 4. In b), the vertex is on the chord's side of the center, so the altitude is \(5-3=2\,\text{cm}\). A congruent side is \(\sqrt{2^2+4^2}=2\sqrt{5}\,\text{cm}\).

a) \(4\sqrt{5}\,\text{cm}\) b) \(2\sqrt{5}\,\text{cm}\)

- Identify the locus determined by each condition separately. - Compare distance from a point with perpendicular distance from a line. - Determine how each parallel line meets the circle.

1. The points exactly \(4\,\text{cm}\) from \(P\) form a circle centered at \(P\) with radius \(4\,\text{cm}\). 2. The points exactly \(4\,\text{cm}\) from \(g\) lie on two lines parallel to \(g\), one on each side. 3. Because \(P\) lies on \(g\), each parallel line is exactly one radius from the center. Each line is tangent to the circle at one point. Therefore, there are \(2\) points that satisfy both conditions.

1. A circle centered at \(P\) with radius \(4\,\text{cm}\) 2. Two lines parallel to \(g\), each \(4\,\text{cm}\) from \(g\) 3. \(2\) points

- Find the distance between two points on the same horizontal line. - Recall the relationship between a tangent and a radius at the point of tangency. - A line perpendicular to a horizontal line is vertical. - Locate the point opposite \(P\) across the center.

1. Points \(M\) and \(P\) have the same y-coordinate, so \(MP = 10 - 6 = 4\,\text{cm}\). Thus, \(r = 4\,\text{cm}\). 2. Radius \(\overline{MP}\) is horizontal. A tangent is perpendicular to the radius at the point of tangency, so \(t_1\) is vertical. 3. The point of tangency for the parallel tangent is the point opposite \(P\) on the horizontal diameter. Move \(4\) units left from \(M(6, 4)\) to get \(Q(2, 4)\).

a) \(r = 4\,\text{cm}\) b) \(\overline{MP}\) is horizontal, and a tangent is perpendicular to the radius at the point of tangency, so \(t_1\) is vertical. c) \(Q(2, 4)\)

- Identify the right triangle formed by a radius, the center-to-chord distance, and part of the chord. - The distance from a point to a line segment is measured perpendicularly. - A perpendicular from the center of a circle bisects a chord. - Apply the Pythagorean theorem to one half of the diagram.

1. The perpendicular segment from the center to a chord bisects the chord. Thus, the radius \(r\), distance \(d\), and half-chord \(\frac{s}{2}\) form a right triangle. 2. Apply the Pythagorean theorem: \(13^2=5^2+\left(\frac{s}{2}\right)^2\). 3. Then \(\left(\frac{s}{2}\right)^2=169-25=144\), so \(\frac{s}{2}=12\,\text{cm}\). 4. Therefore, \(s=2\cdot12=24\,\text{cm}\).

The chord length is \(s=24\,\text{cm}\).

- Identify the right angle in the diagram. - A perpendicular from the center of a circle bisects a chord. - The radius is the hypotenuse of the smaller right triangle. - Use half of the chord length in the calculation.

1. The perpendicular from the center to the chord bisects the chord, so half of the chord is \(15\,\text{cm}\). 2. The radius is the hypotenuse of a right triangle whose legs are \(d\) and \(15\,\text{cm}\). 3. Apply the Pythagorean theorem: \(d^2+15^2=17^2\). 4. Then \(d^2=289-225=64\), so \(d=\sqrt{64}=8\,\text{cm}\).

The distance is \(d=8\,\text{cm}\).

- What do the two outermost lines of sight tell you about the apparent size of a round balloon? - What relationship is useful where a line just touches a circle? - How could symmetry help you work with one of the two sightlines? - Which trigonometric ratio relates radius and center distance?

1. The observer's two sightlines are tangent to the balloon's circular cross section. A radius to a tangent point is perpendicular to the sightline. 2. Half the visual angle is \(0.75^\circ\). The radius is opposite this half-angle and the distance \(d\) to the center is the hypotenuse. 3. Thus, \(\sin(0.75^\circ)=\frac{12}{d}\). 4. Solve: \(d=\frac{12}{\sin(0.75^\circ)}\approx916.76\,\text{m}\).

The observer is approximately \(916.76\,\text{m}\) from the balloon's center.

- What theorem relates a radius to a tangent at the point where they meet? - After justifying the two angles at \(B\) and \(C\), consider quadrilateral \(ABMC\). - Use the marked angle in the diagram with the interior-angle sum of a quadrilateral.

1. A radius drawn to a point of tangency is perpendicular to the tangent, so \(m\angle MBA=m\angle MCA=90^\circ\). 2. The diagram gives \(m\angle BAC=44^\circ\). The angle measures in quadrilateral \(ABMC\) sum to \(360^\circ\). 3. Therefore, \(44^\circ+90^\circ+m\angle BMC+90^\circ=360^\circ\). 4. Thus, \(m\angle BMC=360^\circ-224^\circ=136^\circ\).

a) Each angle is formed by a radius and a tangent at the point of tangency, so \(\angle MBA=\angle MCA=90^\circ\). b) \(m\angle BMC=136^\circ\)

- Focus on the two tangent segments that meet at each vertex. - Give the equal tangent lengths temporary variables and express each side as a sum of two of them. - Compare the sums of opposite sides only after deriving the relationship. - Then substitute the three lengths shown in the diagram.

1. Tangent segments from the same external point to a circle have equal lengths. 2. Let the tangent lengths from vertices \(A\), \(B\), \(C\), and \(D\) be \(p\), \(q\), \(r\), and \(s\), respectively. Then \(AB=p+q\), \(BC=q+r\), \(CD=r+s\), and \(DA=s+p\). 3. Therefore, \(AB+CD=(p+q)+(r+s)=(q+r)+(s+p)=BC+DA\). 4. Read the side lengths from the diagram: \(8+10=11+d\). 5. Thus \(18=11+d\), so \(d=7\).

a) Tangent segments from the same external point are equal. b) \(AB+CD=BC+DA\) c) \(d=7\)

- What are the angle measures in an equilateral triangle? - What angle is formed by a radius and a tangent at the point of tangency? - Which quadrilateral in the diagram has a known angle sum?

1. Because triangle \(ABP\) is equilateral, \(m\angle APB=60^\circ\). 2. A radius is perpendicular to a tangent at the point of tangency, so \(m\angle MAP=m\angle MBP=90^\circ\). 3. The angle measures of quadrilateral \(AMBP\) sum to \(360^\circ\). 4. Therefore, \(m\angle AMB=360^\circ-90^\circ-90^\circ-60^\circ=120^\circ\).

\(m\angle AMB=120^\circ\)

- Which sides of triangles \(OMA\) and \(OMB\) must match? - How can you establish that the angles next to \(M\) are equal? - What is the sum of two adjacent angles along a straight line?

1. Segments \(OA\) and \(OB\) are radii, so \(OA=OB\). 2. Since \(M\) is the chord's midpoint, \(AM=MB\). Segment \(OM\) belongs to both triangles. 3. Thus \(\triangle OMA\cong\triangle OMB\) by SSS. 4. The congruent angles \(\angle OMA\) and \(\angle OMB\) form a linear pair, so each is \(90^\circ\). Therefore, \(OM\perp AB\).

\(\triangle OMA\cong\triangle OMB\) by SSS. Their adjacent angles at \(M\) are equal and supplementary, so both are right angles and \(OM\perp AB\).

- Translate the perpendicular-bisector condition into a relationship between two chords. - Connect equal chords in one circle to their intercepted arcs. - Use the stated location of \(C\) on the minor arc.

1. Since \(C\) lies on the perpendicular bisector of \(\overline{AB}\), \(CA=CB\). 2. Congruent chords in the same circle intercept congruent minor arcs. 3. Therefore, minor arc \(AC\) is congruent to minor arc \(CB\). 4. Because \(C\) lies on the minor arc \(AB\), these two congruent arcs partition that arc. 5. Thus \(C\) is the midpoint of the minor arc \(AB\).

The perpendicular-bisector condition gives \(CA=CB\). Equal chords intercept equal arcs, so minor arc \(AC\) equals minor arc \(CB\). Therefore, \(C\) is the midpoint of minor arc \(AB\).

- Identify the complete locus of points equidistant from \(A\) and \(B\). - Intersect that locus with the given circle. - Use the perpendicular from the circle's center to the resulting chord.

1. Every point equidistant from \(A\) and \(B\) lies on the perpendicular bisector \(p\) of \(\overline{AB}\). 2. Therefore, the required points are the intersections \(P\) and \(Q\) of \(p\) with the circle. 3. Let \(M\) be the foot of the perpendicular from \(O\) to \(p\). Since \(OM=3\,\text{cm}\), the perpendicular from the center to chord \(\overline{PQ}\) bisects the chord. 4. In right triangle \(OMP\), \(MP=\sqrt{5^2-3^2}=4\,\text{cm}\). 5. Thus \(PQ=2MP=8\,\text{cm}\).

The points are the two intersections \(P\) and \(Q\) of the circle with the perpendicular bisector of \(\overline{AB}\). Their distance is \(8\,\text{cm}\).

- What must be true of the center's distances to both tangents? - Where are points equidistant from two intersecting lines? - How do two adjacent angles around the intersection add up?

1. The center of a circle tangent to both lines has equal perpendicular distances to \(\ell\) and \(m\). 2. Points equidistant from intersecting lines lie on their two angle-bisector lines. 3. Each point of either bisector, except \(O\), is the center of a circle whose positive radius equals that point's distance from either line. 4. Adjacent angles between the original lines total \(180^\circ\). Their bisected angles total \(90^\circ\), so the two bisector lines are perpendicular.

The possible centers are all points on the two angle-bisector lines of \(\ell\) and \(m\), except \(O\). Those lines are perpendicular because their angles are halves of supplementary angles.

- Compare the two right triangles formed by \(\overline{OM}\) and the two halves of the chord. - Which sides in those triangles are already known to be congruent? - After finding half the chord, relate it to the center-to-chord distance and a radius. - For the second chord, compare its length with \(\overline{AB}\) before doing any new calculation.

1. Triangles \(OMA\) and \(OMB\) are right triangles. Also, \(OA=OB\) because they are radii, and \(OM\) is a common leg. 2. By hypotenuse-leg congruence, \(\triangle OMA\cong\triangle OMB\). Therefore, \(AM=MB\), so \(M\) bisects \(\overline{AB}\). 3. Since \(AB=16\,\text{cm}\), each half is \(8\,\text{cm}\). With \(OM=6\,\text{cm}\), the radius is \(OA=\sqrt{6^2+8^2}=10\,\text{cm}\). 4. Chord \(CD\) has the same length as chord \(AB\). Congruent chords in the same circle are equidistant from the center, so the perpendicular distance from \(O\) to \(CD\) is also \(6\,\text{cm}\).

a) \(M\) bisects \(\overline{AB}\), so \(AM=MB=8\,\text{cm}\). b) \(10\,\text{cm}\) c) \(6\,\text{cm}\)

- Distinguish the external part \(PA\) from the whole secant \(PB\). - What length-product relationship connects a tangent and a secant from one external point? - The tangent length is used twice in that relationship.

1. For a tangent and a secant from the same external point, \(PT^2=PA\cdot PB\). 2. The diagram gives \(PT=12\) and \(PA=8\), so \(12^2=8\cdot PB\). 3. Thus \(144=8\cdot PB\), so \(PB=18\).

\(PB=18\)

- On each secant, distinguish the segment outside the circle from the entire segment from \(P\) to the farther intersection point. - The upper whole secant includes both labeled pieces on that ray. - Express the lower whole secant using the unknown \(CD\) before writing the product relationship. - Check that the two products use the same kind of pair: outside length and whole-secant length.

1. From the diagram, \(PA=4\,\text{cm}\) and \(AB=5\,\text{cm}\), so the whole upper secant is \(PB=4+5=9\,\text{cm}\). 2. Let \(CD=x\,\text{cm}\). Since \(PC=3\,\text{cm}\), the whole lower secant is \(PD=3+x\,\text{cm}\). 3. For two secants from the same external point, external segment times whole secant is equal: \(PA\cdot PB=PC\cdot PD\). 4. Thus \(4\cdot9=3(3+x)\), so \(36=9+3x\) and \(x=9\). 5. Multiplying the outside segment by only the inside segment is incorrect because the theorem uses the distance from \(P\) all the way to the farther intersection point on each secant.

\(CD=9\,\text{cm}\). The theorem uses external segment \(\times\) whole secant, not external segment \(\times\) inside segment.

- Split each chord into two equal halves. - What lengths are shared by the two right triangles from the center? - Which relationship determines the remaining leg?

1. The perpendicular from a circle's center bisects a chord, so \(AM=\frac{1}{2}AB\) and \(CN=\frac{1}{2}CD\). 2. Equal chords give \(AM=CN\), and both right triangles have hypotenuse equal to radius \(r\). 3. Hence \(OM^2=r^2-AM^2=r^2-CN^2=ON^2\), and both lengths are nonnegative, so \(OM=ON\).

Equal chords have equal half-lengths. The two center-to-chord right triangles have the same radius as hypotenuse, so \(OM=ON\).

- Which triangles can you make using the tangent, the secant, and a chord? - Can you find an angle shared at the external point? - What other angles are linked to the same arc?

1. The angle at \(P\) is common to triangles \(PTA\) and \(PBT\), because \(P,A,B\) are collinear. 2. The tangent-chord theorem makes \(\angle PTA\) equal to \(\angle PBT\). The triangles are similar by AA. 3. Corresponding sides give \(\frac{PT}{PB}=\frac{PA}{PT}\). 4. Therefore, \(PT^2=PA\cdot PB\).

\(\triangle PTA\sim\triangle PBT\) by a shared angle at \(P\) and the tangent-chord theorem. Then \(\frac{PT}{PB}=\frac{PA}{PT}\), so \(PT^2=PA\cdot PB\).

- Which extra chord could create two triangles using the two secants? - Which angle is common at the external point? - Can angles on the circle give you a second equal pair?

1. The triangles share the angle at \(P\), since \(P,A,B\) are collinear and \(P,C,D\) are collinear. 2. Angles \(PAD\) and \(PCB\) are equal because both are supplements of inscribed angles that intercept arc \(BD\). 3. Therefore, \(\triangle PAD\sim\triangle PCB\) by AA. 4. Corresponding sides give \(\frac{PA}{PC}=\frac{PD}{PB}\), so \(PA\cdot PB=PC\cdot PD\).

\(\triangle PAD\sim\triangle PCB\) by AA. The proportion \(\frac{PA}{PC}=\frac{PD}{PB}\) gives \(PA\cdot PB=PC\cdot PD\).

- Split each chord in half before making a right triangle. - Which side of each right triangle is the radius? - Compare the remaining perpendicular distances only after calculating both.

1. A perpendicular from the center bisects each chord. 2. For the \(24\,\text{cm}\) chord, the half-length is \(12\,\text{cm}\), so its distance is \(\sqrt{13^2-12^2}=5\,\text{cm}\). 3. For the \(10\,\text{cm}\) chord, the half-length is \(5\,\text{cm}\), so its distance is \(\sqrt{13^2-5^2}=12\,\text{cm}\). 4. Thus the longer chord is nearer the center; its half-length uses more of the fixed radius.

The \(24\,\text{cm}\) chord is \(5\,\text{cm}\) from the center; the \(10\,\text{cm}\) chord is \(12\,\text{cm}\) from the center. The longer chord is nearer because with the same radius, its longer half-chord leaves a shorter perpendicular leg.

- Use the radius-tangent perpendicularity theorem. - Apply the quadrilateral angle sum. - Identify the two pairs of adjacent congruent sides. - Recall the diagonal property of a kite.

1. Radii \(\overline{MS}\) and \(\overline{MT}\) are perpendicular to the tangent segments at \(S\) and \(T\). Thus, \(\angle MSP = 90^\circ\) and \(\angle MTP = 90^\circ\). 2. The interior angles of quadrilateral \(MSPT\) sum to \(360^\circ\). Therefore, \(124^\circ + 90^\circ + \angle SPT + 90^\circ = 360^\circ\). 3. Solving gives \(\angle SPT = 56^\circ\). 4. Since \(MS = MT\) as radii and \(PS = PT\) as tangent segments from the same external point, \(MSPT\) is a kite. 5. The diagonal joining the vertices where the congruent sides meet, \(\overline{MP}\), is the perpendicular bisector of the other diagonal. Therefore, \(ST \perp MP\).

a) The angles at \(S\) and \(T\) are \(90^\circ\). b) \(\angle SPT = 56^\circ\). c) \(ST \perp MP\). Quadrilateral \(MSPT\) is a kite, and its symmetry diagonal \(\overline{MP}\) is perpendicular to \(\overline{ST}\).

- Convert the basketball's diameter to a radius. - A tangent sight line is perpendicular to the radius at the point of tangency. - Use half of the visual angle in one of the two congruent right triangles.

1. The basketball has radius \(r=\frac{9.5}{2}=4.75\,\text{in.}\). 2. A line of sight tangent to the circular cross section forms a right triangle with the player's eye and the center of the ball. The angle at the eye is half of \(\alpha\). 3. From the diagram, the center-to-eye distance is \(32\,\text{in.}\), so \(\sin\left(\frac{\alpha}{2}\right)=\frac{4.75}{32}\). 4. Thus, \(\alpha=2\sin^{-1}\left(\frac{4.75}{32}\right)\approx 17.07^\circ\). 5. To the nearest tenth, \(\alpha\approx 17.1^\circ\).

The visual angle is approximately \(17.1^\circ\).

- What does tangency tell you about the distances from the circle's center to the two bases? - What does the position of the circle's center tell you about distances to the four sides? - Form right triangles using the radius and half of each base. - Use the trapezoid's height to find a leg.

1. The bases touch the circle at opposite endpoints of a diameter, so their distance apart is \(6\,\text{cm}\). 2. Because the center \(O\) is equidistant from the two sides meeting at each vertex, the segment from a vertex to \(O\) bisects its interior angle. 3. At each end of the shorter base, the interior angle is \(120^\circ\), giving a \(60^\circ\) half-angle. In the right triangle to the tangent point, half the shorter base is \(\frac{3}{\tan(60^\circ)}=\sqrt{3}\,\text{cm}\). 4. At each end of the longer base, the half-angle is \(30^\circ\). Half the longer base is \(\frac{3}{\tan(30^\circ)}=3\sqrt{3}\,\text{cm}\). 5. Thus, the bases are \(2\sqrt{3}\,\text{cm}\) and \(6\sqrt{3}\,\text{cm}\). Each leg spans a height of \(6\,\text{cm}\), so its length is \(\frac{6}{\sin(60^\circ)}=4\sqrt{3}\,\text{cm}\).

Bases: \(2\sqrt{3}\,\text{cm}\) and \(6\sqrt{3}\,\text{cm}\). Each leg: \(4\sqrt{3}\,\text{cm}\).

- What angle pair comes from the two chords crossing at \(P\)? - How could connecting other points on the circle form two triangles with those angles? - What property of inscribed angles could establish a second equal pair?

1. Consider the triangles formed by joining \(A\) to \(C\) and \(D\) to \(B\). 2. Angles \(APC\) and \(DPB\) are vertical angles, so they are equal. 3. Angles \(PAC\) and \(PDB\) intercept the same arc \(BC\), so they are equal. Thus triangles \(APC\) and \(DPB\) are similar by AA. 4. Corresponding sides give \(\frac{PA}{PD}=\frac{PC}{PB}\), so \(PA\cdot PB=PC\cdot PD\).

Add chords \(AC\) and \(BD\). Then \(\angle APC=\angle DPB\) as vertical angles, and \(\angle PAC=\angle PDB\) because both intercept arc \(BC\). Thus \(\triangle APC\sim\triangle DPB\), so \(\frac{PA}{PD}=\frac{PC}{PB}\) and \(PA\cdot PB=PC\cdot PD\).

- Which four points can lie on a second circle because of the altitudes? - Relate an angle at \(F\) to the angle at \(C\). - Which angle does the tangent-chord theorem connect to \(AB\)?

1. Altitudes give \(\angle BEC=\angle BFC=90^\circ\), so \(B,C,E,F\) lie on the circle with diameter \(BC\). 2. Opposite angles in that cyclic quadrilateral satisfy \(m\angle EFB+m\angle ECB=180^\circ\). 3. Since \(A,F,B\) are collinear, \(m\angle EFA+m\angle EFB=180^\circ\). Hence \(\angle EFA=\angle ECB\). 4. Since \(A,E,C\) are collinear, \(\angle ECB=\angle ACB\). 5. By the tangent-chord theorem, the angle between \(t\) and \(AB\) equals \(\angle ACB\). Thus \(EF\) and \(t\) make equal corresponding angles with \(AB\), so they are parallel.

The right angles make \(BCEF\) cyclic. Its opposite angles show \(\angle EFA=\angle ACB\), the same angle that \(t\) makes with \(AB\). Therefore, \(EF\parallel t\).