Posted by Jane on Monday, January 30, 2012 at 4:41am.
Find an equation of the circle tangent to x + y = 3 at (2,1) and with center on 3x  2y  6 = 0

Solid mensuration  Damon, Monday, January 30, 2012 at 9:21am
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 = x1
the center is also on 3 x  2 y = 6
so
3 x  2 (x1) = 6
x = 4
y = 3
center at (4,3)
so form is
(x4)^2 + (y3)^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 = (13)^2 + (24)^2
r^2 = 4 + 4 = 8
so
(x4)^2 + (y3)^2 = 8
check my arithmetic, I did that fast.