Circle Theorems
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Fri Apr 04 2025 12:42:25 GMT+0000 (Coordinated Universal Time).
These are the solutions to the WASSCE past questions on Circle Theorems.
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Length of Arc, Area of Sector, Area of Circle, Circumference of Circle
Except stated otherwise, use:
$
d = diameter \\[3ex]
r = radius \\[3ex]
L = arc\:\:length \\[3ex]
A = area\;\;of\;\;sector \\[3ex]
\theta = central\:\:angle \\[3ex]
\pi = \dfrac{22}{7} \\[5ex]
RAD = radians \\[3ex]
^\circ = DEG = degrees \\[7ex]
\underline{\theta\;\;in\;\;DEG} \\[3ex]
L = \dfrac{2\pi r\theta}{360} \\[5ex]
\theta = \dfrac{180L}{\pi r} \\[5ex]
r = \dfrac{180L}{\pi \theta} \\[5ex]
A = \dfrac{\pi r^2\theta}{360} \\[5ex]
\theta = \dfrac{360A}{\pi r^2} \\[5ex]
r = \dfrac{360A}{\pi\theta} \\[7ex]
\underline{\theta\;\;in\;\;RAD} \\[3ex]
L = r\theta \\[5ex]
\theta = \dfrac{L}{r} \\[5ex]
r = \dfrac{L}{\theta} \\[5ex]
A = \dfrac{r^2\theta}{2} \\[5ex]
\theta = \dfrac{2A}{r^2} \\[5ex]
r = \sqrt{\dfrac{2A}{\theta}} \\[7ex]
Circumference\:\:of\:\:a\:\:circle = 2\pi r = \pi d \\[3ex]
L = \dfrac{2A}{r} \\[5ex]
r = \dfrac{2A}{L} \\[5ex]
A = \dfrac{Lr}{2}
$
Circle Theorems
(1.) The angle in a semicircle is a right angle (an angle of 90°).
(2.) Angles in the same segment of a circle are equal.
OR
Angles subtended by a chord of a circle in the same segment of the circle are equal.
(3.) The angle which an arc of a circle subtends at the center is twice the angle which the same
arc of the circle subtends at the circumference.
OR
The measure of any angle inscribed in a circle is half the measure of the intercepted arc.
(4.) The sum of the interior opposite angles of a cyclic quadrilateral is 180°
OR
The interior opposite angles of a cyclic quadrilateral are supplementary
(5.) The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
(6.) The radius of a circle is perpendicular to the tangent of the circle at the point of contact.
This implies that the angle between the radius of a circle and the tangent to the circle at the point of
contact is 90°
(7.) Intersecting Tangents Theorem or Intersecting Tangent-Tangent Theorem and
Angle of Intersecting Tangents Theorem
If two tangents are drawn from the same external point:
(a.) the two tangents are equal in length
(b.) the line joining the external point and the centre of the circle bisects the angle formed by the two
tangents.
(c.) the line joining the external point and the centre of the circle bisects the angle formed by the two
radii.
(8.) Alternate Segment Theorem
The angle between a tangent to a circle and a chord drawn from the point of contact, is equal to the angle in
the alternate segment.
(9.) If a line drawn from the center of the circle bisects a chord, then:
(a.) it bisects its arc (the angle opposite the chord) and
(b.) it is perpendicular to the chord.
(10.) If a line drawn from the center of the circle is perpendicular to a chord, then:
(a.) it bisects the chord and
(b.) it bisects its arc (the angle opposite the chord).
(11.) Intersecting Chords Theorem
When two chords intersect, the product of the lengths of the segments of one chord is equal to the product of
the
lengths of the segments of the other chord.
(12.) Angle of Intersecting Chords Theorem
The angle formed when two chords intersect is equal to half the sum of the intercepted arcs.
(13.) Intersecting Secants Theorem or Intersecting Secant-Secant Theorem
In the intersection of two secants from the same external point:
the product of the distance between the first point and the external point and the distance between the
second point and the external point for the first secant is equal to
the product of the distance between the first point and the external point and the distance between the
second point and the external point for the second secant.
(14.) Angle of Intersecting Secants (Inside the Circle) Theorem
The angle formed when two secants intersect inside a circle is equal to half the sum of the intercepted arcs.
(15.) Angle of Intersecting Secants (Outside the Circle) Theorem
The angle formed when two secants intersect outside a circle is equal to half the difference of the
intercepted arcs.
(16.) Intersecting Secant-Tangent Theorem or Intersecting Tangent-Secant Theorem
In the intersection of a secant and a tangent from the same external point:
the product of the distance between the first point and the external point and the distance between the
second point and the external point for the secant is equal to
the square of the distance between the point of contant and the external point for the tangent.
(17.) Angle of Intersecting Secant-Tangent Theorem
The angle formed when a secant and a tangent intersect outside a circle is equal to half the difference of the
intercepted arcs.
In the diagram, P, Q, R and S are points on the circle centre K.
$\overline{KR}$ is a bisector of $\angle SRQ$
$\angle KSP$ = 41° and $\angle SKR$ = 80°
Find:
(a.) $\angle RQP$
(b.) $\angle SPQ$
$ \underline{\triangle SKR} \\[3ex] \angle KSR = \angle KRS = x^\circ ...\text{base angles of isosceles triangle SKR} \\[3ex] x^\circ + x^\circ + 80^\circ = 180^\circ ...\text{sum of angles in a triangle} \\[3ex] 2x = 180 - 80 \\[3ex] 2x = 100 \\[3ex] x = \dfrac{100}{2} \\[5ex] x = 50 \\[3ex] \therefore \angle KSR = \angle KRS = 50^\circ \\[5ex] Also: \\[3ex] \angle RSP = \angle KSR + \angle KSP ...diagram \\[3ex] \angle RSP = 50 + 41 = 91^\circ \\[5ex] \angle KRQ = \angle KSR = 50^\circ ...\overline{KR} \text{ is a bisector of } \angle SRQ \\[3ex] \angle SRQ = \angle KRS + \angle KRQ ...diagram \\[3ex] \angle SRQ = 50 + 50 = 100^\circ \\[5ex] (a.) \\[3ex] \angle RQP + \angle RSP = 180^\circ ...\text{interior opposite angles of a cyclic quad.} \\[3ex] \angle RQP = 180 - \angle RSP \\[3ex] = 180 - 91 \\[3ex] = 89^\circ \\[5ex] (b.) \\[3ex] \angle SPQ + \angle SRQ = 180^\circ ...\text{interior opposite angles of a cyclic quad.} \\[3ex] \angle SPQ = 180 - \angle SRQ \\[3ex] = 180 - 100 \\[3ex] = 80^\circ $
If $|\overline{PQ}| = |\overline{QR}| = 17\;cm \;\;|\overline{PR}| = 16\;cm$ and M is the midpoint of $|\overline{PR}|$, calculate:
(a.) $|\overline{QM}|$
(b.) correct to the nearest whole number, the radius of the circle.
Let us represent the information on a diagram
Please note the colors
$ (a.) \\[3ex] Let\;\;|\overline{QM}| = h \\[3ex] \underline{\triangle QMR} \\[3ex] h^2 + 8^2 = 17^2 ...\text{Pythagorean Theorem} \\[3ex] h^2 = 17^2 - 8^2 \\[3ex] h^2 = 289 - 64 \\[3ex] h^2 = 225 \\[3ex] h = \sqrt{225} \\[3ex] h = 15\;cm \\[5ex] $ We can solve (b.) using at least two approaches.
Use any approach you prefer.
Let us update our diagram
Let O be the centre of the circle
Let r be the radius of the circle
Please note the colors
$ (b.) \\[3ex] \underline{\text{1st Approach: Pythagorean Theorem}} \\[3ex] \underline{\triangle OMR} \\[3ex] r^2 = (15 - r)^2 + 8^2 ...\text{Pythagorean Theorem} \\[3ex] r^2 = (15 - r)(15 - r) + 64 \\[3ex] r^2 = 225 - 15r - 15r + r^2 + 64 \\[3ex] r^2 = 225 - 30r + r^2 + 64 \\[3ex] r^2 - r^2 + 30r = 225 + 64 \\[3ex] 30r = 289 \\[3ex] r = \dfrac{289}{30} \\[5ex] r = 9.633333333 \\[3ex] r \approx 10\;cm ...\text{to the nearest whole number} \\[3ex] $ Let:
a, b, c be the sides of the triangle
s be the perimeter A be the area of the triangle
$ \underline{\text{2nd Approach: Heron's Formula and Circumradius Formula}} \\[3ex] a = 16\;cm \\[3ex] b = 17\;cm \\[3ex] c = 17\;cm \\[3ex] s = \dfrac{a + b + c}{3} = \dfrac{16 + 17 + 17}{2} = 25 \\[5ex] s - a = 25 - 16 = 9 \\[3ex] s - b = 25 - 17 = 8 \\[3ex] s - c = 25 - 17 = 8 \\[3ex] s(s - a)(s - b)(s - c) = 25(9)(8)(8) = 14400 \\[3ex] A = \sqrt{s(s - a)(s - b)(s - c)}...\text{Heron's Formula} \\[3ex] A = \sqrt{14400} \\[3ex] A = 120\;cm^2 \\[5ex] r = \dfrac{abc}{4A} \\[5ex] r = \dfrac{16 \cdot 17 \cdot 17}{4 \cdot 120} \\[5ex] r = 9.633333333 \\[3ex] r \approx 10\;cm ...\text{to the nearest whole number} $
