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 $$PC_2 = \sqrt{64 + 4r^{2}} = 2\sqrt{16 + r^{2}}$$$$PC_2 - PC_1 = 3r$$$$\Rightarrow 2\sqrt{16 + r^{2}} - \sqrt{16 + r^{2}} = 3r$$$$\Rightarrow \sqrt{16 + r^{2}} = 3r$$$$\Rightarrow 16 + r^{2} = 9r^{2}$$$$\Rightarrow r^{2} = 2$$Maths

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