The equation of a circle is (x - 3)^2 + (y + 2)^2 = 25. The point (8, -2) is on the circle. What is the equation of the line that is tangent to the circle at (8, -2)?

Question

misty1212 · Answer

can you use calculus or do you have to use geometry?

solomonzelman · Answer

if calc.

~  take the derivative:
~  solve for the dy/dx  (or for y' , whatever your notation is)
~  plug in the point (8-,2) into the derivative.  (this will be the slope)
~  use point slope formula

solomonzelman · Answer

excuse me, plug in point (8,-2)

mathstudent55 · Answer

No calculus approach:
1. Find the slope of the radius the passes through (8, -2)
2. The tangent is the line that is perpendicular to that radius and passes through (8,-2).

solomonzelman · Answer

I said, if it is calculus...

anonymous · Answer

I think mathstudent55 is just outlining what to do in the other case.  @SolomonZelman gave one explanation using calculus, and @mathstudent55 gave the other without it.

solomonzelman · Answer

oh, I see.. can very well be that...

solomonzelman · Answer

anyways, it depends on what teacher requires... my teacher told me to show the derivative of ln(x) using lim[h->0] f(x+h) ... that rule. and I had go through 3 limit changes.

anonymous · Answer

This is geometry by the way, sorry for not clarifying that

misty1212 · Answer

you have the equation of the circle given, so you know the center right?

misty1212 · Answer

center is $(3,-2)$ find the slope of the line from $(3,-2)$ to $(8,-2)$ which you really don't need any work to find, it is horizontal
the tangent line will be vertical

misty1212 · Answer

did you get this?

anonymous · Answer

Yes! Thank you!

misty1212 · Answer

vertical line through $(8,-2)$is all