Question

$P(a,b)$ is a point in the first quadrant. If two circles which pass through $P$ and touch both the coordinate axes and cuts each other at right angles, then

• A. $a^2-6ab+b^2=0$
• B. $a^2+2ab-b^2=0$
• C. $a^2-4ab+b^2=0$
• D. $a^2-8ab+b^2=0$

Answer 1

The correct option is C $a^2-4ab+b^2=0$

Any circle touching both the axes is given by $x^2+y^2-2rx–2ry+r^2=0$

Given the circle passes through $P(a,b)$.

$⟹a^2+b^2-2r(a+b)+r^2=0$

This is a quadratic equation in terms of $r$.

Let $r_1,r_2$ be the solutions of the equations.

$⟹r_1+r_2=\frac{-b}{a}=\frac{2(a+b)}{1}–\left(1\right)$

and $(r_1.r_2)=\frac{c}{a}=\frac{a^2+b^2}{1}–\left(2\right)$

Given that the two circles intersect orthogonally.

$⟹C_1{C}_2^2=r_1^2+r_2^2$

$⟹(r_1–r_2{)}^2+(r_1–r_2{)}^2=r_1^2+r_2^2$

$⟹r_1^2+r_2^2=4r_1{r}_2$

$⟹(r_1+r_2{)}^2=6r_1{r}_2$

Using (1) and (2)

$\left(2\right(a+b){)}^2=6(a^2+b^2)$

$⟹4(a^2+b^2+2ab)=6a^2+6b^2$

$⟹a^2+b^2–4ab=0$