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## Homework Statement

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}##

## Homework Equations

Quadratic discriminant, sum and product of roots

## 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. :(

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