Question 397119: Please help me solve this question!
"Determine the equation of the tangent to the circle (x-2)^2+(y+3)^2=13 at the point (5,-5)"
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! Right away, we can tell that the center of the circle is at (2, -3).
It suffices to find the slope of the line between (2, -3) and (5, -5). This is because the radius of a circle is perpendicular to the line tangent to the circle at that point (5, -5).
The slope of the line is  . Therefore the slope of the line tangent to the circle is 3/2 (since these two lines are perpendicular). Since it runs through the point (5, -5) we can find the y-intercept:
-5 = 5(3/2) + b, where b is the y-intercept. Solving, we get b = -25/2, so the equation is  .
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