Posted by **Marrion** on Saturday, September 26, 2009 at 4:00am.

A circle has the equation x^2+y^2-4x+6y-12=0

and i have already found that the centre has a coordinate of (2,-3), and a radius of 5 units. But i need to find points on the circle that has the gradient of -4/3 when a tangent is drawn on the point. How do i do that???

- Gradient of circle? -
**MathMate**, Saturday, September 26, 2009 at 7:27pm
Assuming you have not done Calculus, we will work without it.

Let O = centre point of the circle

If the tangent to the point P has a gradient of m, what can you say about the gradient (slope) of the radius OP, denoted by p?

We know that OP and the tangent are perpendicular to each other, so m*p=-1.

You can therefore find p knowing m.

What is the line passing through the centre (2,-3) with a slope of p?

(y-(-3)) = p(x-2)

The intersection of this line with the circle will give you the required tangent points. Note that there are two such points. It will be the solution of a quadratic equation.

- Gradient of circle? -
**Marrion**, Sunday, September 27, 2009 at 2:13am
I see! Thanks!

- Gradient of circle? :) -
**MathMate**, Sunday, September 27, 2009 at 8:16am
You're welcome!

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