posted by ceasar1 .
Find the equation of the smaller circle tangent to both axes and passing the point (-3,6)
If the circle is tangent to both axes and passes through (-3,6), I see its centre to be in the 2nd quadrant.
If the radius is r, then its centre must be (-r,r) and its equation is
(x+r)^2 + (y-r)^2 = r^2
x^2 + 2rx + r^2 + y^2 - 2ry + r^2 = r^2
9 - 6r + r^2 + 36 - 12r = 0
r^2 - 18r + 45 = 0
(r-3)(r-15) = 0
r = 3 or r = 15
since you want the smaller circle ... r = 3
(x+3)^2 + (y-3)^2 = 9