Circles: math test solutions, Grade 10 – download the PDF

1 Tangents / 3 pts

(a) \(OT \perp PT\), so \(PT^2 = 15^2 - 9^2 = 144\) and \(PT = 12\) cm. (2 pts)

(b) Tangent segments from the same external point are congruent, so \(PS = 12\) cm. (1 pt)

2 Inscribed angles / 4 pts

(a) \(\angle ACB\) intercepts \(\widehat{AB}\): \(84 \div 2 = 42^\circ\). (1 pt)

(b) \(\angle ABD\) intercepts \(\widehat{DA}\): \(110 \div 2 = 55^\circ\). (1 pt)

(c) \(\angle ABC\) intercepts \(\widehat{CD} + \widehat{DA} = 180^\circ\): \(m\angle ABC = 90^\circ\), so \(AC\) is a diameter. (1 pt)

(d) \(m\angle DAB = \tfrac{1}{2}(96 + 70) = 83^\circ\) and \(m\angle BCD = \tfrac{1}{2}(110 + 84) = 97^\circ\). \(83 + 97 = 180\), so they are supplementary. (1 pt)

3 Arc and sector / 4 pts

(a) \(\ell = \dfrac{48}{360}\cdot 2\pi\cdot 15 = 4\pi \approx 12.57\) cm. (2 pts)

(b) \(A = \dfrac{48}{360}\cdot \pi\cdot 225 = 30\pi \approx 94.25\) cm\(^2\). (2 pts)

4 Segment lengths / 4 pts

(a) \(8\cdot 3 = 4\cdot ED\), so \(ED = 6\). (2 pts)

(b) \(6^2 = 4\cdot w\), so \(w = 9\). (1 pt) The chord is \(9 - 4 = 5\). (1 pt)

5 Equation of a circle / 3 pts

(a) \((x^2 - 12x + 36) + (y^2 + 4y + 4) = -15 + 36 + 4\), so \((x - 6)^2 + (y + 2)^2 = 25\). (1 pt)

(b) Center \((6, -2)\), radius \(5\). (1 pt)

(c) \((2 - 6)^2 + (1 + 2)^2 = 16 + 9 = 25\), so yes, the point lies on the circle. (1 pt)

6 Angle outside a circle / 2 pts

The angle is half the difference of the arcs: \(\tfrac{1}{2}(130 - 46)\) (1 pt) \(= 42^\circ\). (1 pt)