Review question

# Can we find the tangents to a circle from a point? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6804

## Solution

Prove that the line $y=mx+c$ will touch the circle $x^2+y^2=25$ if $c^2=25(1+m^2).$

For the line to touch the circle there must be exactly one point of intersection between them—the line must be a tangent to the circle.

To find the points of intersection of the line and circle we need to solve their equations for $x$ and $y$.

Substituting $y = mx + c$ into $x^2 + y^2 = 25$ gives \begin{align*} &x^2 + (mx+c)^2 = 25\\ \iff\quad& x^2 + m^2x^2 + 2mcx + c^2 = 25\\ \iff\quad& (1+m^2)x^2 + 2mcx + c^2 - 25 = 0 \end{align*}

This is a quadratic equation in $x$, and so to have exactly one point of intersection, we require the discriminant to be zero. The discriminant here is $(2mc)^2 - 4 (1+m^2) (c^2-25)= 4 (25 (1+m^2) - c^2),$ where we have expanded, rearranged and taken out a factor of $4$.

So if $c^2 = 25(1+m^2)$ then the discriminant is $0$. Therefore the line and circle intersect at exactly one point, and so the line touches the circle.

We can, in fact, prove more than this; we also have that the line touches the circle if $c^2=25(1+m^2)$.

We were asked to prove that if $c^2=25(1+m^2)$, then the line touches the circle; in fact, we see that $c^2=25(1+m^2)$ if and only if the line touches the circle.

Hence or otherwise find the equations of the two tangents to this circle from the point $(2,11)$.

We can save time by using the first part of the question to solve the next part.

We want to find straight lines of the form $y=mx+c$ which go through the point $(2,11)$ and are tangent to the circle.

Substituting the coordinates $(2,11)$ into $y=mx+c$ gives $11 = 2m + c$, and from the first part we must have $c^2 = 25(1+m^2)$.

Now we have simultaneous equations for $m$ and $c$ which we can solve.

Substituting for $c$ gives $(11-2m)^2 = 25(1+m^2),$ which rearranges to give $21m^2 + 44m - 96 = 0.$ This can be factorised to give $(3m-4) (7m+24) = 0.$ So $m = \dfrac{4}{3}$ or $m = - \dfrac{24}{7}$.

If $m = \dfrac{4}{3}$, then $c = 11 - 2\left(\dfrac{4}{3}\right) = \dfrac{25}{3}$, and if $m = - \dfrac{24}{7}$ then $c = 11 - 2 \left(-\dfrac{24}{7}\right) = \dfrac{125}{7}$.

So the two tangents are $y= \dfrac{4}{3} x + \dfrac{25}{3}$ and $y= - \dfrac{24}{7} x + \dfrac{125}{7}.$

We could write these equations in the form $ay+bx+c = 0$, giving $3y-4x-25=0$ and $7y+24x-125=0.$