What must the radius of C be for PR to be tangent to it at Q?
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Answer:
D. 8
Step-by-step explanation:
Draw segment CQ which will be perpendicular to tangent PR.
So, triangle CQR will be right angled triangle right angle at Q.
Let radius of the circle be r.
Therefore,
CQ = CS = r
By Pythagoras Theorem:
[tex] {(r + 9)}^{2} = {r}^{2} + {15}^{2} \\ \\ {r}^{2} + 18r + 81 = {r}^{2} + 225 \\ \\ 18r = 225 - 81 \\ \\ 18r = 144 \\ \\ r = \frac{144}{18} \\ \\ r = 8[/tex]