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Question

two circles are externally tangent. Lines PAB and PA'B' are common tangents with A and A' on the smaller circle and B and B' on the larger circle. If PA = AB = 4, then the square of radius of circle is


Solution


$$\dfrac{AC_1} {PA} = \dfrac{BC_2} {PB} \Rightarrow BC_2 = 2AC_1$$
$$PC_1 = \sqrt{16 + r^{2}}$$
and $$PC_2 = \sqrt{64 + 4r^{2}} = 2\sqrt{16 + r^{2}}$$
$$PC_2 - PC_1 = 3r$$
$$\Rightarrow 2\sqrt{16 + r^{2}} - \sqrt{16 + r^{2}}  = 3r$$
$$\Rightarrow \sqrt{16 + r^{2}}  = 3r$$
$$\Rightarrow 16 + r^{2} = 9r^{2}$$
$$\Rightarrow r^{2} = 2$$




1620385_1766658_ans_11a7357caa7e4e968aa172d84157b3f2.png

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