SOLUTION: Find the equation of the circle that passes through the point (-3,-4) and touches line x-y+7=0 at (-5,2). I got x^2+y^2+15x+2y+7+k(3x+y+13) and took the mid points of this equatio

Question 948801: Find the equation of the circle that passes through the point (-3,-4) and touches line x-y+7=0 at (-5,2).
I got x^2+y^2+15x+2y+7+k(3x+y+13) and took the mid points of this equation and substituted to the derieved equation of perpendicular of the given tangent to get the value of k = -11/4. But this doesn't work. What's wrong?
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
Examine some of the data carefully. The given line is equivalent to y=x+7. You will find that the given point (-5,2) is on this line. Knowing that should be helpful for you.

(-5,2) is a point of tangency. The line perpendicular to y=x+7 which contains (-5,2) will also contain the center point of the circle. m=-1 for this line. y=-x-3 is the line which contains the center point of the circle. You may well be able to continue from here.
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