This answer key gives a complete worked solution for each problem on the circles worksheet. Check each step against your work, especially which arc an angle intercepts, and compare your figures with the diagrams shown for the harder problems.
Problem 1: Similar circles
(a) Translate A so its center lands on the center of B. Then dilate about that center by the scale factor \(\frac{10}{4}=\frac{5}{2}=2.5\).
(b) \(\frac{2\pi(10)}{2\pi(4)}=\frac{20\pi}{8\pi}=2.5\).
(a) Translation, then dilation with scale factor 2.5. (b) The ratio is 2.5.
Problem 2: Ratio of circumferences
\(C_P=2\pi(3)=6\pi\) and \(C_Q=2\pi(12)=24\pi\). The ratio is \(\frac{24\pi}{6\pi}=4\), the same as the ratio of the radii, \(\frac{12}{3}=4\).
\(C_P=6\pi\) in, \(C_Q=24\pi\) in, ratio 4.
Problem 3: Central angle and arc
A central angle equals its intercepted arc, so minor arc \(AB\) is \(72^{\circ}\). The full circle is \(360^{\circ}\), so major arc \(AB\) is \(360^{\circ}-72^{\circ}=288^{\circ}\).
Minor arc \(72^{\circ}\); major arc \(288^{\circ}\).
Problem 4: Inscribed angle from an arc
An inscribed angle is half its intercepted arc: \(\frac{1}{2}(110^{\circ})=55^{\circ}\).
\(m\angle ACB=55^{\circ}\)
Problem 5: Arc from an inscribed angle
The arc is twice the inscribed angle: \(2(38^{\circ})=76^{\circ}\). A central angle equals the arc.
Arc \(76^{\circ}\); central angle \(76^{\circ}\).
Problem 6: Inscribed angle algebra
The angle is half the arc: \(3x+5=\frac{1}{2}(8x-10)\), so \(3x+5=4x-5\) and \(x=10\).
Angle: \(3(10)+5=35^{\circ}\). Arc: \(8(10)-10=70^{\circ}\). Check: \(\frac{70}{2}=35\).
\(x=10\), angle \(35^{\circ}\), arc \(70^{\circ}\).
Problem 7: Triangle in a semicircle
\(\angle ACB\) is inscribed in a semicircle, so it is \(90^{\circ}\). The angles of a triangle sum to \(180^{\circ}\): \(\angle B=180^{\circ}-90^{\circ}-34^{\circ}=56^{\circ}\).
\(\angle ACB=90^{\circ}\) and \(\angle B=56^{\circ}\).
Problem 8: Semicircle algebra
\(\angle ACB=90^{\circ}\), so the other two angles sum to \(90^{\circ}\): \((2x+4)+(3x+6)=90\), \(5x+10=90\), \(x=16\).
\(2(16)+4=36^{\circ}\) and \(3(16)+6=54^{\circ}\). Check: \(36+54=90\).
\(x=16\); the angles are \(36^{\circ}\) and \(54^{\circ}\).
Problem 9: Angles on the same arc
Both inscribed angles intercept arc \(AB\), so each is half of \(82^{\circ}\): \(\frac{1}{2}(82^{\circ})=41^{\circ}\).
\(m\angle ACB=m\angle ADB=41^{\circ}\)
Problem 10: Radius and tangent
The radius \(OA\) is perpendicular to the tangent, so triangle \(OAP\) has a right angle at \(A\). \(OP=\sqrt{6^{2}+8^{2}}=\sqrt{100}=10\).
\(OP=10\)
Problem 11: Angle at a tangent
Since \(\overline{OA}\perp\ell\), \(\angle OAP=90^{\circ}\). Then \(m\angle APO=180^{\circ}-90^{\circ}-52^{\circ}=38^{\circ}\).
\(m\angle APO=38^{\circ}\)
Problem 12: Two tangents
\(\angle AOB\) is a central angle, so it is \(130^{\circ}\). The radii meet the tangents at right angles, so quadrilateral \(OAPB\) has angles \(90^{\circ}\), \(90^{\circ}\) and \(130^{\circ}\) at \(A\), \(B\) and \(O\). The angles of a quadrilateral sum to \(360^{\circ}\): \(m\angle APB=360^{\circ}-90^{\circ}-90^{\circ}-130^{\circ}=50^{\circ}\).
\(m\angle AOB=130^{\circ}\) and \(m\angle APB=50^{\circ}\).
Problem 13: Equal tangent segments
Triangle \(OAP\) has a right angle at \(A\): \(PA=\sqrt{13^{2}-5^{2}}=\sqrt{144}=12\). Triangle \(OBP\) is congruent, so \(PB=12\).
\(PA=PB=12\)
Problem 14: Distance to a chord
The perpendicular from the center bisects the chord, so half the chord is 12. The distance \(d\) satisfies \(d^{2}+12^{2}=13^{2}\), so \(d=\sqrt{169-144}=5\).
The distance is 5 units.
Problem 15: Chord from a distance
Half the chord is \(\sqrt{17^{2}-8^{2}}=\sqrt{225}=15\), so the chord is \(2(15)=30\).
The chord is 30 units long.
Problem 16: Equal chords
Equal chords cut equal arcs: \(4x+6=6x-14\), so \(20=2x\) and \(x=10\).
Arc \(AB\): \(4(10)+6=46^{\circ}\). Arc \(CD\): \(6(10)-14=46^{\circ}\).
\(x=10\); each arc is \(46^{\circ}\).
Problem 17: Inscribed quadrilateral
Opposite angles of an inscribed quadrilateral are supplementary. \(m\angle C=180^{\circ}-105^{\circ}=75^{\circ}\) and \(m\angle D=180^{\circ}-82^{\circ}=98^{\circ}\). Check: \(105+82+75+98=360\).
\(m\angle C=75^{\circ}\) and \(m\angle D=98^{\circ}\).
Problem 18: Inscribed quadrilateral algebra
Opposite angles sum to \(180^{\circ}\): \((4x+10)+(2x+20)=180\), \(6x+30=180\), \(x=25\).
\(\angle A=4(25)+10=110^{\circ}\) and \(\angle C=2(25)+20=70^{\circ}\).
\(x=25\), \(\angle A=110^{\circ}\), \(\angle C=70^{\circ}\).
Problem 19: Ferris wheel seats
(a) \(\frac{360^{\circ}}{12}=30^{\circ}\).
(b) Three gaps: \(3(30^{\circ})=90^{\circ}\).
(c) Her angle is inscribed and intercepts the \(90^{\circ}\) arc, so it is \(\frac{1}{2}(90^{\circ})=45^{\circ}\).
(a) \(30^{\circ}\). (b) \(90^{\circ}\). (c) \(45^{\circ}\).
Problem 20: Pizza cut
The radius is 10 inches and half the chord is 6 inches. The distance from the center is \(\sqrt{10^{2}-6^{2}}=\sqrt{64}=8\) inches.
The cut is 8 inches from the center.
Problem 21: Plaza lights
Angle \(ALB\) is inscribed in a semicircle, so \(m\angle ALB=90^{\circ}\). Triangle \(ALB\) is right with hypotenuse 80: \(LB=\sqrt{80^{2}-48^{2}}=\sqrt{6400-2304}=\sqrt{4096}=64\).
\(m\angle ALB=90^{\circ}\); Liam is 64 feet from pole \(B\).
Problem 22: Pond path
The path is tangent, so \(OT\perp TP\). \(OP=\sqrt{30^{2}+40^{2}}=50\) yards. The nearest point of the water is on segment \(OP\), \(30\) yards from \(O\), so Mei is \(50-30=20\) yards from the water.
\(OP=50\) yards; the shortest distance to the water is 20 yards.
Problem 23: Central and inscribed algebra
The central angle is twice the inscribed angle: \(12x-20=2(5x-5)\), so \(12x-20=10x-10\), \(2x=10\) and \(x=5\).
\(\angle AOB=12(5)-20=40^{\circ}\) and \(\angle ACB=5(5)-5=20^{\circ}\).
\(x=5\), \(\angle AOB=40^{\circ}\), \(\angle ACB=20^{\circ}\).
Problem 24: Tangents and arcs algebra
The angle at \(P\) and the minor arc are supplementary: \((4x-2)+(6x+2)=180\), so \(10x=180\) and \(x=18\).
\(\angle P=4(18)-2=70^{\circ}\) and arc \(AB=6(18)+2=110^{\circ}\).
\(x=18\), \(\angle P=70^{\circ}\), arc \(AB\) \(=110^{\circ}\).
See every chapter on the Geometry math page.