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The basic equation for a straight line is $y = mx + b$, where $b$ is the height of the line at $x = 0$ and $m$ is the gradient. The basic equation for a circle is $(x - c)^2 + (y - d)^2 = r^2$, where $r$ is the radius and $c$ and $d$ are the $x$ and $y$ shifts of the center of the circle away from $(0,\ 0)$.

I'm trying to come up with an equation for determining the intersection points for a straight line through a circle. I've started by substituting the "y" value in the circle equation with the straight line equation, seeing as at the intersection points, the y values of both equations must be identical. This is my work so far:

$$(x - c)^2 + (mx + b - d)^2 = r^2$$ $$x^2 + c^2 - 2xc + m^2x^2 + (b - d)^2 + 2mx(b - d) = r^2$$ I'll shift all the constants to one side $$x^2 - 2xc + m^2x^2 + 2mx(b - d) = r^2 - c^2 - (b - d)^2$$ $$(m^2 - 1)x^2 + 2x(m(b - d) - c) = r^2 - c^2 - (b - d)^2$$

That's as far as I can get. From what I've gathered so far, I ought to be able to break down the left side of this equation into a set of double brackets like so: $$(ex + f)(gx + h)$$ where $e,\ f,\ g$ and $h$ are all constants. Then I simply have to solve each bracket for a result of $0$, and I have my $x$ coordinates for the intersections of my two equations. Unfortunately, I can't figure out how to break this equation down.

Any help would be appreciated.

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Why don't you just solve for $x$? You know the constans, right? – M.B. Nov 4 '12 at 12:22
I'm trying to develop an equation that will deduce what x is for any set of constants, not just a specific set. – Cambot Nov 4 '12 at 12:41

Let's say you have the line $y = mx + c$ and the circle $(x-p)^2 + (y-q)^2 = r^2$.

First, substitute $y = mx + c$ into $(x-p)^2 + (y-q)^2 = r^2$ to give

$$(x-p)^2 + (mx+c-q)^2 = r^2 \, . $$

Next, expand out both brackets, bring the $r^2$ over to the left, and collect like terms:

$$(m^2+1)x^2 + 2(mc-mq-p)x + (q^2-r^2+p^2-2cq+c^2) = 0 \, .$$

This is a quadratic in $x$ and can be solved using the quadratic formula. Let us relabel the coefficients to give $Ax^2 + Bx + C = 0$, then we have

$$x = \frac{-B \pm \sqrt{B^2-4AC}}{2A} \, . $$

If $B^2-4AC < 0$ then the line misses the circle. If $B^2-4AC=0$ then the line is tangent to the circle. If $B^2-4AC > 0$ then the line meets the circle in two distinct points.

Since $y=mx+c$, we can see that if $x$ is as above then

$$y = m\left(\frac{-B \pm \sqrt{B^2-4AC}}{2A}\right) + c \, . $$

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Ah! Thank you! This certainly looks like it'll do what I'm trying to do! – Cambot Nov 4 '12 at 13:20
Be careful to manage vertical line !!! – Eric Ouellet May 13 at 19:26

You've got a quadratic in $x$. Use the discriminant to see if it has real solutions, and if so how many. Then solve the quadratic for $x$, and substitute the solution(s), if any, back into the equation for the line.

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Have you checked the general solution on wolframalpha? You've already made your quadratic equation and so, you can get the roots of $x$ as the solutions that can then be used to get the values of $y$.

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