https://i.stack.imgur.com/krUqK.png
Example we have one line and one circle above, and its equation are:
Circle: $(x-a)^2 + (y-b)^2 = r^2$
Line: $y = m(x-x_1) + y_1$
How do we find the intersection coordinate?
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1$\begingroup$ Substitute your equation for $y$ (from the line) into the equation of the circle, multiply everything out, and solve the quadratic equation to get $x$. Use the equation of the line to get a $y$ for each $x$. $\endgroup$– Christopher Carl HeckmanMar 13, 2016 at 23:58
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1 Answer
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Well, why not bash here? Solve the equations in terms of $x,$ as follows:
Given: $$(x - a)^{2} + (y - b)^{2} = r^{2}$$ $$y = m\left(x - x_{1}\right) + y_{1}$$
With substitution: $$(x - a)^{2} + \left(m\left(x - x_{1}\right) + \left(y_{1} - b\right)\right)^{2} = r^{2}$$ $$\left(x^{2} - 2ax + a^{2}\right) + m^{2}\left(x^{2} - 2x_{1}x + x_{1}^{2}\right) + 2m\left(x - x_{1}\right)\left(y_{1} - b\right) + \left(y_{1}^{2} - 2by_{1} + b^{2}\right) = r^{2}.$$
I leave you to use the quadratic equation on $x$ and finish the problem. Note that sometimes, there will be no solution.
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1$\begingroup$ The distance between the circle centre and the intersection point is: $$D=\left|\frac{\sqrt{-4(m^2+1)((-mx_1+y_1-b_1)^2+a_1^2-c^2)+(2m(-mx_1+y_1-b_1)-2a_1)^2}-2m(-mx_1+y_1-b_1)+2a_1)}{2m^2+2}\right|$$ Bravo sir, thank you :) $\endgroup$ Mar 14, 2016 at 0:26
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$\begingroup$ No problem! When in doubt, try bashing the problem ;) $\endgroup$– K. JiangMar 14, 2016 at 0:43