Q.FIND THE EQUATION OF THE CIRCLE TOUCHING THE BOTH THE COORDINATE AXES AND PASSING THROUGH THE POINT (3,-6)?

general equation of circle is

(x-h)² + (y-k)² = R²

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

(x-h)² + (y-h)² = h²

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

(3-h)² + (-6-h)² = h²

 h² +6h+45 = 0

 descriminant = -ve , hence h is imaginary so there will be no circle under this condition....

Let the radius of the circle be r. Then as the circle touches both the coordinate axes, the coordinates of its centre are (r,r)

and its distance from the point (3,-6) is the radius of the circle

(r-3)^2+(r+6)^2=r^2

now calculate it