You have an error in your problem statement. In order for the given line to be tangent to a circle at a particular point, that point must be an element of the solution set of the equation of the line. Note that
However, since
I'm going to assume that you just made a typo and that the actual point of tangency is (4, -4).
First determine the slope of the given line. The easiest way is just to put the equation into slope intercept form and then examine the coefficient on
Note that a radius at a point of tangency is perpendicular to the tangent line. Perpendicular lines have slopes that are negative reciprocals.
Use the point-slope form of an equation of a line to write an equation of the line containing the radius at (4,-4).
The center of the circle lies on the line whose equation we just derived. Further, the radius of the circle is the distance from this line to the other given point, i.e. the distance from the given point to the point of intersection with the derived line and a perpendicular to that line through the given point (-3,-5).
A line perpendicular to the just derived line has a slope identical to the given tangent line, namely
Use the point-slope form again:
Now we have an equation for a line containing a radius with endpoints at the point (-3,-5) and the center of the circle.
The center is the point of intersection between
and
This system lends itself tidily to solution by the substitution method (just set the two RHSs equal)
Which is true if and only if
Consequently, our center is at (0,-1)
The measure of the radius can be found by using the distance formula on the derived center and either of the two given points. I will leave it as an exercise for the student to verify that the radius in this case is indeed 5.
The equation for a circle centered at
So our equation is:
Simplifying:
John
My calculator said it, I believe it, that settles it
