Posted by John on .
What is the equation of a circle passing through (4,4) and tangent to the line 2x3y+9=0 at (3,1)?

Analytic Geometry 
Steve,
The slope of the given line is 2/3
So, the radius through (3,1) is perpendicular, with slope 3/2.
So, the line through the center and (3,1) is
y1 = 3/2 (x+3)
Now, the line through the two points forms a chord of the circle, and its slope is 5. So, the radius perpendicular to that point has slope 1/5 and passes through the midpoint of the chord at (7/2,3/2).
y+3/2 = 1/5 (x+7/2)
So, now we have two lines which intersect at (1,2), the center of the circle. So, the circle is
(x+1)^2 + (y+2)^2 = r^2
The distance from (1,2) to (4,4) or (3,1) is √13, so our circle is
(x+1)^2 + (y+2)^2 = 13