# how to find the tangent-lines of a circle, given eq. of the circle and point of the tangents outside the circle?

So I have a circle: $x^2 + y^2 = 25$ and a point $P = (7,1)$ of the tangent-lines and I have to find the equations of these tangent-lines.

So I know that the radius is $5$ and that the center is $C = (0,0)$. I also know that the equation of a tangent-line is $y=mx+h$.

I can substitute $x$ and $y$ of $P$ in the equation of the tangent-line, so I would have $x$ and $y$ but how do I find $h$ and $m$ then?

• Mar 4, 2018 at 12:50

The lines that contains $P$ have equations: $y-1=m(x-7)$ and we want the lines that have only a common point with the circle $x^2+y^2=25$.
The common points are the solutions of the system $$\begin{cases} x^2+y^2=25\\ y-1=m(x-7) \end{cases}$$ Substituting $y$ ( or $x$) from the second to the first equation you find a second degree equation in $x$ (or $y$) and this equation has only one solution (only one common point) iff its discriminant is null.
Note that the discriminant $\Delta(m)$ is a second degree polynomial in $m$ so The equation $\Delta(m)=0$ can have $2$ distinct solutions ( if the point $P$ is external to the circle and we have two tangent lines), one solution ( if the circle contains $P$ so that we have only one tangent line) or no real solutions if the point $P$ is inside the circle.
Anyway, solving $\Delta(m)=0$ you find the slope of the tangents lines, if they exist.
HINT: write $$y=mx+n$$ for the searched Tangent line, then we have the equation $$y=m(x-7)+1$$ plug this in the equation of the given circle then you will get $$x^2+(m(x-7)+1)^2=25$$ solve this equation for $x$ and set the discriminant equal to Zero you will get $$-24m^2+14m+24=0$$ to compute $m$