Question

What is the equation of the line that is tangent to the circle at (8, -2)?

Answer 1

Is that all the information?. I think you would need to know at least the center or radius of the circle.

Answer 2

I was also given- The equation of a circle is (x - 3)2 + (y + 2)2 = 25. The point (8, -2) is on the circle.

Answer 3

I'm sorry, but I won't be of much help as I'm new to Calculus myself, but I would solve for y, that will give you a y = +- sqrt(....., then find the derivative y (y'). That gives you the the equation for the slop of the tangent line. Then when you have that, insert it into the the point slop formula
(y - y1) = m(x-x1)

Where you replace m with y', and then solve for y, this gives you y = mx +c, which is the equation of the tangent line at point (8,-2)

Answer 4

Its not calculus lol its geometry, but it's ok :)

Answer 5

ahhh - sorry. Can't be of any help then

Answer 6

that's ok, thank you anyway :)

Answer 7

well - actually, you have the center at 3, -2. And r = 5 (r^2 = 25), so you have a tangent line which is vertical, and the slop is undefined.

Answer 8

right, point 8,-2 is to the right of the center. The tangent line is perpendicular to the line from center. Since the lince from center is horizontal, the tangetn must be vertical.
I never took geometry, so I might not use the proper terms.

Answer 9

so the equation of the tangent line is x = 8

Answer 10

but you better double check :)

Answer 11

You're right, I just plug that in and submitted and got it right. Thanks!!!