Question

Two points P and Q are taken on the line joining the points A(0, 0) and B(3a, 0) such that AP = PQ = QB. Circles are drawn on AP, PQ and QB as diameters. The locus of the point S, the sum of the squares of the tangents from which to the three circles is equal to $b^2$, is

• A. $x^{2}+y^{2}-3ax+2a^{2}-b^{2} = 0$
• B. $3\left(x^{2}+y^{2}\right)-9ax +8a^{2}-b^{2} = 0$
• C. $x^{2}+y^{2}-5ax +6a^{2}-b^{2} = 0$
• D. $x^{2}+y^{2}-ax-b^{2} = 0$

Answer 1

Answer: (B)

Points P and Q are $(a, 0)$ and $(2a, 0)$
So, the circles are
$\left(x-0\right)\left(x-a\right) + \left(y-0\right)\left(y-0\right) = 0 \Rightarrow x^{2} + y^{2} - ax = 0$
$\left(x-a\right)\left(x-2a\right) + \left(y-0\right)\left(y-0\right) = 0 \Rightarrow x^{2} + y^{2} - 3ax + 2a^{2} = 0$
$\left(x-2a\right)\left(x-3a\right) + \left(y-0\right)\left(y-0\right) = 0$
$\Rightarrow x^{2}+ y^{2}-5ax+6a^{2}= 0$
Now, S be $\left(h, k\right)$ then
$h^{2}+k^{2}-ah+h^{2}+k^{2}-3ah+2a^{2}+h^{2}+k^{2} -5ah +6a^{2} = b^{2}$
$\Rightarrow 3\left(h^{2}+k^{2}\right) -9ah +8a^{2} = b^{2}$