mathematics
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 + (yr)^2 = r^2
x^2 + 2rx + r^2 + y^2  2ry + r^2 = r^2
at (3,6)
9  6r + r^2 + 36  12r = 0
r^2  18r + 45 = 0
(r3)(r15) = 0
r = 3 or r = 15
since you want the smaller circle ... r = 3
equation:
(x+3)^2 + (y3)^2 = 9