## How to find tangent's equation? (circle & parabola)

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

### How to find tangent's equation? (circle & parabola)

Hi,

I have a parabola y = x^2 and

a circle x^2 + y^2 - 5/2y + 9/16 = 0

Parabola and circle have a common tangent in the point with the x-coordinate of sqrt(3)/2.

I need to find the tangent's equation.

So far, I found the circles' center (0, 5/4) and radius = 1

Any help appreciated.
mitnord

Posts: 13
Joined: Mon Feb 23, 2009 11:17 pm

mitnord wrote:I have a parabola y = x^2 and a circle x^2 + y^2 - 5/2y + 9/16 = 0

Parabola and circle have a common tangent in the point with the x-coordinate of sqrt(3)/2.

I need to find the tangent's equation.

So far, I found the circles' center (0, 5/4) and radius = 1

You are given the x-coordinate of the shared point of tangency. Plug this into either one of the original equations to find the y-coordinate, and thus the coordinates of that point.

For a line to be "tangent" to another figure, it must just touch the figure, rather that crossing it. For a circle, this means that the tangent line at a point must be perpendicular to the radius line at that point. (Otherwise, the line would pass through the circle, rather than only touching it.)

You have the center and the point of tangency. What is the slope between these two points? What then is the perpendicular slope?

Plug this perpendicular slope $m$ and the point of tangency $(x_1,\, y_1)$ into the point-slope formula $y\, -\, y_1\, =\, m(x\, -\, x_1)$ to find your line equation.

If you get stuck, please reply showing how far you have gotten. Thank you!

stapel_eliz

Posts: 1703
Joined: Mon Dec 08, 2008 4:22 pm