Question

Let $P(a,b)$ be a point in the first quadrant. Circles are drawn through $P$ touching the coordinate axes such that the length of common chord for the smaller circle is maximum. If possible values of $\frac{a}{b}$ is $k_1$ and $k_2,$ then $k_1+k_2$ is equal to

Answer 1

Let $r_1,r_2$ $(r_1<r_2)$ be the radii of the circles.
Since the two circles touch the coordinate axes,
Equations will be of the form $x^2+y^2-2r(x+y)+r^2=0$
Since it passes through $P\equiv (a,b),$
$\therefore a^2+b^2-2r(a+b)+r^2=0$
which is a quadratic equation in $r.$
$\Rightarrow r_1=a+b-\sqrt{2ab},r_2=a+b+\sqrt{2ab}$

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 the smaller circle,
i.e., $(a+b-\sqrt{2ab},a+b-\sqrt{2ab})$ should lie on the common chord.
Putting the coordinates in chord equation, we get $a+b=2\sqrt{2ab}$
$\Rightarrow a^2+b^2=6ab$
Dividing LHS and RHS by $b^2,$
${\left(\frac{a}{b}\right)}^2+1=6\left(\frac{a}{b}\right)$
$\Rightarrow {\left(\frac{a}{b}\right)}^2-6\left(\frac{a}{b}\right)+1=0$
So, from sum of the roots of quadratic equation
$k_1+k_2=6$