the point A (-3,-7) lies on the circle centre (5,1) find the equation of the tangent to the circle at a?

Question

please help! i h8 geometry lol thanks

i dont know the equation of the line between the two points. all the questions gives is what i posted.

Answer 1

Whoops - below is a calculus way to do it - you're proabbly not read for that yet ;)

I'm assuming that the center of the circle is (5,1) and the point of tangency we want is at (-3, -7)

The radius of our circle is Sqrt(8^2 + 8^2) = 8 Root(2) by the distance formula.

The equation for our circle is then:

(y-1)^2 + (x-5)^2 = 128

Using implicit differentiation:

2(y-1)dy/dx + 2(x-5) = 0

The point is (-3, -7), so

2(-7-1)dyd/dx + 2(-3-5) = 0
-16dy/dx = 16

dy/dx = -1 (the slope)

You probably could have figured that out without calculus after noticing that the horizontal distance and vertical distance were both the same to the point on the circle.

Answer 2

Well, what's the equation of the line between the two points given?

The line you're solving for is perpendicular to that other line, at the point A. I think you can just assume this latter fact.

And I hope that, given the equation for one line, you know how to find the equation for its perpendicular at a given point.