# Tangent to a circle

1. Mar 5, 2014

### sooyong94

1. The problem statement, all variables and given/known data
Given that the line $y=mx+c$ is a tangent to the circle $(x-a)^{2} +(y-b)^{2} =r^{2}$, show that $(1+m^{2}) r^{2}=(c-b+ma)^{2}$

2. Relevant equations
Quadratic discriminant, sum and product of roots

3. The attempt at a solution
I substituted y=mx+c into the equation of the circle, and this is what I have:
$(x-a)^{2} +(mx+c-b)^{2} =r^{2}$
Then I expanded and simplified into this:
$(1+m^{2})x^{2}+(2mc-2mb-2a)x+a^{2}+b^{2}-2bc+c^{2}-r^{2}=0$

Now if I have to use the quadratic discriminant it would be tedious to work with, unfortunately. :(

Last edited by a moderator: Mar 20, 2014
2. Mar 5, 2014

### scurty

There's a lot of Algebraic manipulation in this problem.

First hint: Find m, which is the same as $\frac{dy}{dx}$
Second hint: solve for c in terms of only a, b, x, y
Third hint: Show both sides of the equations are equal to each other (again, only in terms of a, b, x, y).

Edit: This is in the Precalculus section. Have you learned how to find $\frac{dy}{dx}$ yet?

Last edited: Mar 5, 2014
3. Mar 20, 2014

### utkarshakash

Use the concept that the radius of circle is equal to the distance of the given line from the centre of circle. You already know the formula for finding the distance of any point from a line and in this case it is simply the centre of circle (a,b). Equate both and you're done.

4. Mar 20, 2014

### pasmith

You should have expanded only so far as
$$(1 + m^2)x^2 + 2(m(c-b) - a)x + a^2 + (c-b)^2 - r^2 = 0$$
Keeping (c-b) as (c-b) throughout will make your calculations much less tedious.