Find an equation of the circle tangent to x + y = 3 at (2,1) and with center on 3x - 2y - 6 = 0

Answer 1

y = - x + 3 so
slope of tangent = -1
slope of line from (2,1) to circle center = -1/-1 = 1 because radius of circle is perpendicular to tangent at intersection
so find the radius line to the tangent point (2,1)
y = +1 x + b
1 = 2 + b
b = -1
so that radius line is y = x-1
the center is also on 3 x - 2 y = 6
so
3 x - 2 (x-1) = 6
x = 4
y = 3
center at (4,3)
so form is
(x-4)^2 + (y-3)^2 = r^2
we still need r^2 but we know center at (4,3) and tangent point at (2,1)
distance between is r
r^2 = (1-3)^2 + (2-4)^2
r^2 = 4 + 4 = 8
so
(x-4)^2 + (y-3)^2 = 8
check my arithmetic, I did that fast.