A circle is inscribed in a semicircle with centre O and diameter AB.
The centre of the circle is the point P, wherePA = PO.
What is the ratio of the radius of the circle to the radius of the semicircle?
A. 4 : 9 B. 3 : 8 C. 3 : 7 D. 2 : 5 E. 1 : 2
B. 3 : 8
We let s be the radius of the semicircle and r be the radius of the circle. We let S and T be the points where the semicircle touches the circle, as shown.
Because the diameter AB is a tangent to the circle at S, ∠PSA = ∠PSO = 90°.
Therefore PSA and PSO are right-angled triangles in which PA = PO and the side PS is common. Hence the triangles PSA and PSO are congruent (RHS).
It follows that OS = ASSince OA is a radius of the semicircle, OA = s.
Therefore OS =12OA =12s.Because the circle touches the semicircle at T, they have a common tangent at T.
The radii PT and OT are perpendicular to this common tangent. Therefore the points O, P and T are collinear.
Hence OP = OT − PT = s − r.
PS is a radius of the circle.
Therefore PS = r. We can now apply Pythagoras’ Theorem to the right-angled triangle PSO to give
r2 + (1/2s)2 = (s − r)2.
Expanding, we obtain
r2 +1/4s2 = s2 − 2rs + r2.
Hence 2rs =3/4 s2.
Since s is not equal 0, we can deduce that r =3s/8. Therefore r : s = 3 : 8.
Be up-to-date with our recent updates, new problems and answers!
Ask about our courses and offerings, and we will help you choose what works best for you.
