Can we find the tangents to a circle from a point?

Can we find the tangents to a circle from a point?

Question

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

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

Suggestion

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

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The blue circle here is $x^2 + y^2 = 25,$ while the green line is $y = mx + c$.

When a line and a circle touch, what can we say about their equations?

What else can we write down about $c$ and $m$ if we know a point on the line?

Solution

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.

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 $7y+24x-125=0.$