SOLUTION: Find an equation of the line tangent to the circle at point P. Sketch the circle and the tangent line in the same coordinate system. 4.) x² + y² - 4x + 6y - 12 = 0; P(5,

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find an equation of the line tangent to the circle at point P. Sketch the circle and the tangent line in the same coordinate system. 4.) x² + y² - 4x + 6y - 12 = 0; P(5,      Log On


   

Question 1164229: Find an equation of the line tangent to the circle at point P. Sketch the
circle and the tangent line in the same coordinate system.
4.) x² + y² - 4x + 6y - 12 = 0; P(5,1)
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


By looking at the equation, or by formally completing the square, we can see that the center of the circle is at (2,-3).

Since the point (5,1) is on the circle, the radius to the point of tangency has a slope of 4/3.

That means the tangent line at that point has a slope of -3/4.

Now you have a slope and a point for the tangent line; use your favorite method to find the equation of the tangent.