Question

Obtain the equation of the circle that touches both the axes and passes through the point $(-6,3)$.

Answer 1

Equation of the circle is

$(x-a{)}^2+(y-a{)}^2=a^2$

It passes through $(-6,3)$

$\therefore (-6-a{)}^2+(3-a{)}^2=a^2\Rightarrow 36+a^2+12a+9+a^2-6a=a^{@}\Rightarrow a^2+6a+45=0$

Here $D<0$ therefore it is an imaginary circle.