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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
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B
222
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C
221
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D
1
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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 $$+ r + RP$$
$$\sqrt{2} + 1 = r (1 + \sqrt{2}) \Rightarrow r = 1$$
108482_116906_ans_2e42b2338dec4640aabf77b8fe7ac769.png

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