How to find the coordinates of the intersection of a line and a circle?

Question

I'm trying to find the coordinates of the intersection of a line and a circle. Here is my depiction

Blue line is horizontal, circle centre $(50,50)$, radius $50$.

Can anyone please help me to find the solution?

Answer 1

We have a circle with centre $(50,50)$ and radius $50$, so equation of the circle is $$(x-50)^2+(y-50)^2=2500$$

We know the line is a horizontal line starting at the point $(100,150)$, and so it has the equation $$y=150$$

We want to find all possible co-ordinates of intersection, so we substitute $y=150$ into the equation of the circle:

\begin{align}(x-50)^2+(150-50)^2&=2500\\ (x-50)^2+100^2&=2500\\ (x-50)^2+10000&=2500\\ (x-50)^2&=-7500\\ x-50&=\pm\sqrt{-7500}\\ x&=50\pm\sqrt{-7500}\end{align}

As this is an imaginary number, the line and the circle do not intersect.

Answer 2

To find the coordinates of the points that are intersection of the curve $$ f(x,y)=0\qquad\qquad(\star) $$ and the line $$ y=mx-b\qquad\qquad(\star\star) $$ just plug in the $y$ from $(\star\star)$ into $(\star)$, i.e. solve the equation $$ f(x,mx+b)=0\qquad\qquad(\star\star\star) $$ for $x$. When the curve is a circle, or just any conic, the latter equation $(\star\star\star)$ is actually an equation of degree $2$ whose solution are easily computable from the very well known quadratic formula. The solutions of $(\star\star\star)$ are the $x$-coordinates of the intersection points. Now you can plug in these values back into $(\star\star)$ to get the $y$-coordinates.