Formulae for calculating line and circle intersection

Question

Very simple problem, yet I can't find a solution online and can't seem to keep track of the algebra.

Given $ax+by+c=0$ and $x^2+y^2=r^2$ , Solve for x

constraints: ($a,b,c =$ any real number) ($x,y,r$ any real number) (there can obviously be $0,1,2$ solutions for $x$)

You have to plug it in and simplify it and end up doing quadratic formula, but I got lost after about a page of algebra. Now I'm pretty frustrated, please help me!

Thanks ya'll

Answer 1

I would first "solve" the linar equation $$y= \frac{c-ax}{b}$$

Now we replace in the quadratic:

$$x^2 + (\frac{c-ax}{b})^2 = r^2$$

Compute that square: $$x^2 + \frac{c^2}{b^2} - 2\frac{c-ax}{b^2} + \frac{a^2}{b^2}x^2 = r^2$$

Or, similarly $$x^2 + \frac{c^2}{b^2} - 2\frac{c}{b^2}- 2\frac{a}{b^2}x + \frac{a^2}{b^2}x^2 = r^2$$

You can now group terms and solve the quadratic for $x$

Answer 2

Sometimes it helps if we try to eliminate fractions.

$$\begin{align} ax+by+c &= 0\\ by &= -ax-c\\ by &= -(ax+c)\\ b^2y^2 &= (ax+c)^2\\ b^2y^2 &= a^2x^2+2acx+c^2\\ \end{align}$$

$$\begin{align} x^2+y^2&=r^2\\ y^2&=r^2-x^2\\ b^2y^2&=b^2r^2-b^2x^2\\ \end{align}$$

Can you take it from here?