Circles: practice solutions, Grade 10 – download the PDF

1 Parts of a circle ★★★

(a) \(r = d \div 2 = 14 \div 2 = 7\) ft.

(b) The longest chord of a circle is a diameter, so it is \(14\) ft.

(c) A line that meets the circle at exactly one point is a tangent.

2 A tangent and a radius ★★★

The radius \(OT\) is perpendicular to the tangent, so triangle \(OTP\) is right-angled at \(T\).

\(PT^2 = OP^2 - OT^2 = 625 - 49 = 576\), so \(PT = 24\) cm.

3 Central and inscribed angle ★★★

(a) A central angle equals its intercepted arc: \(112^\circ\).

(b) An inscribed angle is half the arc: \(112^\circ \div 2 = 56^\circ\).

4 Angle in a semicircle ★★★

An angle inscribed in a semicircle is a right angle, so \(m\angle C = 90^\circ\).

The angles of a triangle add up to \(180^\circ\), so \(m\angle B = 180 - 90 - 37 = 53^\circ\).

5 Length of an arc ★★★

A \(60^\circ\) arc is \(\dfrac{60}{360} = \dfrac{1}{6}\) of the circle.

\(\ell = \dfrac{1}{6}\cdot 2\pi \cdot 6 = 2\pi \approx 6.28\) cm.

6 Area of a sector ★★★

\(A = \dfrac{90}{360}\cdot \pi \cdot 10^2 = \dfrac{1}{4}\cdot 100\pi = 25\pi \approx 78.5\) ft\(^2\).

The sprinkler waters about \(78.5\) square feet.

7 Writing an equation ★★★

\((x - (-1))^2 + (y - 4)^2 = 3^2\), that is \((x + 1)^2 + (y - 4)^2 = 9\).

For \((2, 4)\): \((2 + 1)^2 + (4 - 4)^2 = 9 + 0 = 9\). The equation is satisfied, so the point lies on the circle.

8 Distance to a chord ★★★

The perpendicular from the center bisects the chord, so a right triangle has hypotenuse \(13\) and one leg \(12\).

\(d^2 = 13^2 - 12^2 = 169 - 144 = 25\), so \(d = 5\) in.

9 Two tangent segments ★★★

Tangent segments from the same external point are congruent: \(3x + 2 = 5x - 10\).

Then \(12 = 2x\), so \(x = 6\). \(PA = 3(6) + 2 = 20\) and \(PB = 5(6) - 10 = 20\). Each segment is \(20\) units long.

10 Triangle inscribed in a circle ★★★

Each angle is half the arc it intercepts (the arc opposite to it).

\(m\angle C = \tfrac{1}{2}(100) = 50^\circ\), \(m\angle A = \tfrac{1}{2}(130) = 65^\circ\), \(m\angle B = \tfrac{1}{2}(130) = 65^\circ\).

Check: \(50 + 65 + 65 = 180\). Angles \(A\) and \(B\) are equal, so the triangle is isosceles with \(CA = CB\).

11 Pizza slices ★★★

The radius is \(8\) in and each slice has a central angle of \(360 \div 8 = 45^\circ\).

(a) \(A = \dfrac{45}{360}\cdot \pi \cdot 64 = 8\pi \approx 25.1\) in\(^2\).

(b) \(\ell = \dfrac{45}{360}\cdot 2\pi \cdot 8 = 2\pi \approx 6.3\) in.

12 Angles from secants and tangents ★★★

(a) Outside the circle: \(\tfrac{1}{2}(110 - 40) = 35^\circ\).

(b) The vertex is on the circle, so the angle is half the intercepted arc: \(\tfrac{1}{2}(148) = 74^\circ\).

13 Intersecting chords ★★★

\(AE\cdot EB = CE\cdot ED\), so \(6\cdot 4 = 3\cdot ED\).

\(24 = 3\cdot ED\), so \(ED = 8\).

14 Circle from a diameter ★★★

The center is the midpoint: \(\left(\dfrac{-2 + 6}{2}, \dfrac{3 + 9}{2}\right) = (2, 6)\).

\(AB = \sqrt{8^2 + 6^2} = 10\), so \(r = 5\).

Equation: \((x - 2)^2 + (y - 6)^2 = 25\).

15 Ferris wheel ★★★

\(\ell = \dfrac{150}{360}\cdot 2\pi \cdot 30 = \dfrac{5}{12}\cdot 60\pi = 25\pi \approx 78.5\) ft.

In meters: \(78.54 \times 0.3048 \approx 23.9\) m.

16 Completing the square ★★★

Complete the square: \((x^2 + 8x + 16) + (y^2 - 10y + 25) = -16 + 16 + 25\).

\((x + 4)^2 + (y - 5)^2 = 25\). It is a circle with center \((-4, 5)\) and radius \(5\).

17 Tangent line to a circle ★★★

The radius \(OT\) has slope \(\dfrac{4}{3}\). The tangent is perpendicular to it, so its slope is \(-\dfrac{3}{4}\).

\(y - 4 = -\dfrac{3}{4}(x - 3)\). Multiplying by \(4\): \(4y - 16 = -3x + 9\), so \(3x + 4y = 25\).

18 Circle through three points ★★★

The angle at \(O\) is a right angle, so the hypotenuse \(AB\) is a diameter of the circle through the three points.

Center: midpoint of \(AB\), \((4, 3)\). Diameter: \(AB = \sqrt{64 + 36} = 10\), so \(r = 5\).

Equation: \((x - 4)^2 + (y - 3)^2 = 25\). Check with \(O\): \(16 + 9 = 25\).

19 Circles of a right triangle ★★★

The hypotenuse is \(\sqrt{81 + 144} = 15\) cm, and it is a diameter of the circumscribed circle, so \(R = 7.5\) cm.

Area \(= \tfrac{1}{2}\cdot 9 \cdot 12 = 54\) cm\(^2\); semiperimeter \(s = \dfrac{9 + 12 + 15}{2} = 18\) cm. Then \(r = \dfrac{54}{18} = 3\) cm.

20 The reef problem ★★★

The first path is a tangent of length \(t = 30\); the second is a secant with external part \(e = 20\).

\(t^2 = w\cdot e\), so \(900 = 20\,w\) and \(w = 45\) m.

The part inside the reef is \(45 - 20 = 25\) m.

21 Area of a segment ★★★

Sector area: \(\dfrac{90}{360}\cdot \pi \cdot 64 = 16\pi \approx 50.27\) cm\(^2\).

Triangle \(AOB\) is right-angled with legs \(8\) and \(8\): area \(= \tfrac{1}{2}\cdot 64 = 32\) cm\(^2\).

Segment area \(= 16\pi - 32 \approx 18.3\) cm\(^2\).

22 Cyclic quadrilateral ★★★

Opposite angles are supplementary. \((2x + 10) + (4x - 40) = 180\), so \(6x - 30 = 180\) and \(x = 35\): \(m\angle A = 80^\circ\), \(m\angle C = 100^\circ\).

\((3y + 5) + (5y - 65) = 180\), so \(8y - 60 = 180\) and \(y = 30\): \(m\angle B = 95^\circ\), \(m\angle D = 85^\circ\).

Check: \(80 + 100 + 95 + 85 = 360\).

23 Hexagon in a circle ★★★

The hexagon is made of \(6\) equilateral triangles of side \(10\). Each has area \(\dfrac{\sqrt{3}}{4}\cdot 100\), so the hexagon has area \(150\sqrt{3} \approx 259.8\) in\(^2\).

Disk: \(\pi \cdot 10^2 \approx 314.2\) in\(^2\).

Difference: \(314.16 - 259.81 \approx 54.4\) in\(^2\).