SOLUTION: Write an equation of the line tangent to the circle given by (x^2 + y^2 - 6x - 4y + 8 = 0) at the point (5,3) in slope intercept form.

Algebra ->  Circles -> SOLUTION: Write an equation of the line tangent to the circle given by (x^2 + y^2 - 6x - 4y + 8 = 0) at the point (5,3) in slope intercept form.      Log On


   

Question 817055: Write an equation of the line tangent to the circle given by (x^2 + y^2 - 6x - 4y + 8 = 0) at the point (5,3) in slope intercept form.
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Completing the Squares for the circle gives its standard form as %28x-3%29%5E2%2B%28y-2%29%5E2=5.
The center point of this circle is then (3,2).

The point on the circle, (5,3) and the center of circle, (3,2) form a slope of 1%2F2, one half. The tangent line to point (5,3) must be -2%2F1=-2.

This tangent line, starting with point-slope form, is highlight%28y-3=-2%28x-5%29%29. Simplify it and put into slope intercept form.