Question

Find the equation to the circle which touches both axes and passes through the point ( - 2, - 3).

Answer 1

Let the centre be $(a,a)$ and radius be $a$

The equation of circle will be

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

The circle passes through $(-2,-3)$

So, the point satisfies the equation of circle

$(-2-a{)}^2+(-3-a{)}^2=a^2$

$a^2+10a+13=0$

$a=-5\pm 4\sqrt{3}$

Thus, the equation of circle will be

$x^2+y^2+2(5\pm \sqrt{12})(x+y)+37\pm 10\sqrt{12}=0$