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 = 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.

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