Question-42 P is a point (a, b) in the first quadrant. If the two cirdes which pass through P and touch both the coordinate axes, cut at right angles, then :
A)
a^{2}-6 a b+b^{2}=0B)
a^{2}+2 a b-b^{2}=0C)
a^{2}-4 a b+b^{2}=0D)
a^{2}-8 a b+b^{2}=0Answer: C)
Explanation
Let the equation of circle which touches both coordinate axes is
(x-r)^{2}+(y-r)^{2}=r^{2} \Rightarrow x^{2}+y^{2}-2 r x-2 r y+r^{2}=0where, \boldsymbol{r} is radius.
It is passing through (a, b), so
r^{2}-2(a+b) r+\left(a^{2}+b^{2}\right)=0Let the roots of this equation are r_{1}, r_{2}
\section*{Sum of roots}
r_{1}+r_{2}=2(a+b)Product of roots
r_{1} \cdot r_{2}=a^{2}+b^{2}Both circles intersect orthogonally each other so
2 r_{1} r_{2}+2 r_{1} r_{2}=r_{1}^{2}+r_{2}^{2} \begin{aligned} & \Rightarrow 6 r_{1} r_{2}=\left(r_{1}+r_{2}\right)^{2} \\ & \Rightarrow 6\left(a^{2}+b^{2}\right)=4(a+b)^{2} \Rightarrow a^{2}-4 a b+b^{2}=0 \end{aligned}