Consider the circles $x^{2} + y^{2} + 3 \sqrt{2} (x + y) = 0$ and $x^{2} + y^{2} + 5 \sqrt{2} (x + y) = 0$ . A third circle is drawn internally tangent to larger circle and externally tangent to smaller circle and tangent to their common diameter. If radius of such circle is $\frac{p}{q}$ (p & q are relatively prime), then value of p – q is

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Detailed Solution

$\begin{matrix} A B = r + 3, B C = 2 \\ A C = 5 - r \\ C D = \sqrt{{(5 - r)}^{2} - r^{2}} \\ A B^{2} = A D^{2} + B D^{2} \\ {(r + 3)}^{2} = r^{2} + {(2 + \sqrt{5 (5 - 2 r)})}^{2} \\ r^{2} + 6 r + 9 = r^{2} + 4 + 5 (5 - 2 r) + 4 \sqrt{5 (5 - 2 r)} \\ 6 r + 5 = 25 - 10 r + 4 \sqrt{5 (5 - 2 r)} \\ {(4 r - 5)}^{2} = 5 (5 - 2 r) \\ 16 r^{2} + 25 - 40 r = 25 - 10 r \\ 16 r^{2} - 30 r = 0 \\ 2 r (8 r - 15) = 0 \\ r = \frac{15}{8} \\ p - q = 7 \end{matrix}$