Question

# If $$C_1 : x^2 + y^2 = (3 + 2 \sqrt{2})^2$$ be a circle and PA and PB are pair of tangents on $$C_1$$, where P is any point on the director circle of $$C_1$$, then the radius of smallest circle which touch $$C_1$$ externally and also the two tangents $$PA$$ and $$PB$$ is

A
233
B
222
C
221
D
1

Solution

## The correct option is D $$1$$$$AQ = 3 + 2 \sqrt{2}$$          $$(\because given)$$$$PQ = 3 \sqrt{2} + 4$$          $$(\because \angle QPA = \dfrac{\pi}{4})$$Let r be required radius$$\therefore 3\sqrt{2} + 4 = 3 + 2 \sqrt{2} + r + r\sqrt{2}$$            $$(\because \angle RPD = \dfrac{\pi}{4})$$$$PQ$$ $$=$$radius of bigger circle $$+ r + RP$$$$\sqrt{2} + 1 = r (1 + \sqrt{2}) \Rightarrow r = 1$$Maths

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