Question

The equation of circle passing through (3,-6) touching both the axes is

Answer 1

We know the , general equation of circle is

$(x-h{)}^2+(y-k{)}^2=r^2$

when circle touches both the axis then h = k = R

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

now this circle passes through (3,-6) so

($3-h{)}^2+(-6-h{)}^2=h^2$

after solving we get this equation-

$h^2+6^h+45=0$

Discriminant = negative , hence h is imaginary so there will be no circle under this condition.