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