Question

The circle $x^2+y^2-6x-6y+9=0$ is inscribed in a triangle which has two of its sides along the coordinate axes. The locus of the circumcentre of the triangle is $x+y-\frac{b}{a+b}xy+a(x^2+y^2{)}^{\frac{1}{b}}=0$. Find $a+b$

Answer 1

Equation of the given circle $(x-3{)}^2+(y-3{)}^2=9$
Let the equation of the third side $AB$ is $\frac{x}{m}+\frac{y}{n}=1$


Length of the perpendicular from $C=3$
$\therefore \frac{|\frac{3}{m}+\frac{3}{n}-1|}{\sqrt{\frac{1}{m^2}+\frac{1}{n^2}}}=3$
$\because$ origin and C lie on the same side of AB
$\Rightarrow \frac{\frac{3}{m}+\frac{3}{n}-1}{\sqrt{\frac{1}{m^2}+\frac{1}{n^2}}}=-3\frac{3}{m}+\frac{3}{n}-1=-3\sqrt{\frac{1}{m^2}+\frac{1}{n^2}}...\left(1\right)$
$\because \angle AOB=\frac{\pi }{2}$
$\therefore AB$ will be the diameter of the circle which touches $AB$ at $K$
$\therefore$ centre of the circle is $(\frac{m}{2},\frac{n}{2})$
Let the locus of circumcenter be $(h,k)\equiv (\frac{m}{2},\frac{n}{2})$
Now from $\left(1\right),$
$\frac{3}{2h}+\frac{3}{2k}-1=-3\sqrt{\frac{1}{(2h{)}^2}+\frac{1}{(2k{)}^2}}\Rightarrow 3h+3k-2hk=-3\sqrt{h^2+k^2}$
$\therefore$ Locus of the centre $(h,k)$ is
$x+y-\frac{2xy}{3}+\sqrt{x^2+y^2}=0$