# Tangent to the circle at a given point

1. Jul 17, 2014

### skrat

1. The problem statement, all variables and given/known data
I basically have the radius of the circle and its displacement from the origin, so $(x-p)^2+(y-q)^2=r^2$

And now I need to find a tangent to the circle at a given point $(a,b)$. Or at least the slope of the tangent.

How would one do that?

2. Relevant equations

3. The attempt at a solution

If the circle had its center in the origin, than the slope of the tangent would be easily $-\frac a b$ but I am not sure. If the circle is not in the origin, is the slope than $-\frac{a-p}{b-q}$ or is it not?

2. Jul 17, 2014

### Staff: Mentor

Can you differentiate the circle's equation, then isolate dy/dx on one side of the equals sign?

No. Try sketching the points, you'll soon see that's not a tangent.

Always sketch what you are dealing with!

Last edited: Jul 17, 2014
3. Jul 17, 2014

### skrat

$(x-p)^2+(y-q)^2=r^2$

$2(x-p)dx+2(y-q)dy=0$

$\frac{dy}{dx}=-\frac{x-p}{y-g}$

:D Easy as that!

THANKS

4. Jul 17, 2014

### epenguin

No easier but more elementary you could consider the equation of any line through (a, b). Apart from going through (a, b) what other condition does the problem require the line to satisfy?