Question

$P(a,b)$ is a point in the first quadrant. Circles are drawn through $P$ touching the coordinate axes such that the length of common chord of these circles is maximum, if possible values of $a/b$ is $k_1$ and $k_2$, then $k_1+k_2$ is equal to

Answer 1

Let $r_1,r_2$ be the radii of the circles.
Equations will be of the form $x^2+y^2-2r(x+y)+r^2=0$
$\because$ it passes through $P\equiv (a,b)$
$\therefore a^2+b^2-2r(a+b)+r^2=0\Rightarrow r_1,r_2=a+b+\sqrt{2ab},a+b-\sqrt{2ab}$
Now the common chord equation is
$S_1-S_2=0\Rightarrow x+y=a+b$
For maximum length of common chord , common chord becomes diameter of smaller circle i.e., $(a+b-\sqrt{2ab},(a+b-\sqrt{2ab})$ should lie on common chord.
putting the coordinates in chord equation
We get, $a+b=2\sqrt{2ab}\Rightarrow a^2+b^2=6ab\Rightarrow {\left(\frac{a}{b}\right)}^2+\frac{a}{b}-6=0$
So from sum of the roots
$k_1+k_2=6$