Review question

# Given two lines crossing a circle, can we show these lengths are equal? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9551

## Solution

The points $S$, $T$, $U$ and $V$ have coordinates $(s,ms)$, $(t,mt)$, $(u,nu)$ and $(v,nv)$ respectively. The lines $SV$ and $UT$ meet the line $y = 0$ at the points with coordinates $(p,0)$ and $(q,0)$, respectively. Show that $p = \frac{(m - n)sv}{ms - nv},$ and write down a similar expression for $q$.

We begin by finding the equation of the line $SV$.

Its gradient is $M = \frac{ms-nv}{s-v},$ and as it passes through the point $(s, ms)$, the line has equation $y - ms = \frac{ms - nv}{s - v}(x - s)$ (using the general form of the equation of a straight line of gradient $M$ passing through $(x_1,y_1)$, namely $y-y_1=M(x-x_1)$).

This meets the line $y = 0$ at the point $(p,0)$, so $p$ satisfies $-ms = \frac{ms - nv}{s - v}(p - s).$

Rearranging gives the required result for $p$. We have \begin{align*} && p - s &= \frac{ms(v-s)}{ms - nv},\quad \text{assuming ms - nv \neq 0}&&\quad\\ \Longrightarrow\quad&& p &= \frac{s(ms - nv) + ms(v-s)}{ms - nv}\\ \Longrightarrow\quad&& p &= \frac{(m-n)sv}{ms - nv}. \end{align*}

Our argument fails if $s=v$, as then the line $SV$ is vertical. But in this case, $p = s = v$, which agrees with the given formula.

Relabelling $p \mapsto q$, $s \mapsto t$ and $v \mapsto u$ gives the equivalent result for $q$. We have $q = \frac{(m - n)tu}{mt - nu}.$

Given that $S$ and $T$ lie on the circle $x^2 + (y - c)^2 = r^2$, find a quadratic equation satisfied by $s$ and by $t$, and hence determine $st$ and $s + t$ in terms of $m$, $c$ and $r$.

Since $S$ and $T$ lie on the circle, $s$ and $t$ are solutions of the equation (using $z$ as the variable, so as not to confuse ourselves with the equation of the circle itself): \begin{align*} z^2 + (mz - c)^2 = r^2\\ \iff\quad& (1 + m^2)z^2 - 2mcz + (c^2 - r^2) = 0\\ \iff\quad& z^2 - \frac{2mc}{1 + m^2}z + \frac{c^2 - r^2}{1 + m^2} = 0. \end{align*}

Note that we may divide by $1 + m^2$ because it’s positive, and so we are certainly not dividing by $0$.

If the roots of the quadratic equation $z^2 + bz + c = 0$ are $z = s$ and $z = t$, then $s + t = -b$ and $st = c$.

You can check this for yourself by expanding $x^2 + bx + c = (z - s)(z - t)$ and comparing coefficients.

Considering the sum and the product of the roots for our quadratic in $z$, we discover $st = \frac{c^2 - r^2}{1 + m^2}$ and $s + t = \frac{2mc}{1 + m^2}.$

Given that $S$, $T$, $U$ and $V$ lie on the above circle, show that $p + q = 0$.

If $U$ and $V$ lie on the circle, then by interchanging letters we have $uv = \frac{c^2 - r^2}{1 + n^2}$ and $u + v = \frac{2nc}{1 + n^2}.$

Using our expressions for $p$ and $q$ from above, we now have \begin{align*} p + q &= \frac{(m - n)sv}{ms - nv} + \frac{(m-n)tu}{mt - nu}\\ &= \frac{(m - n)}{(ms - nv)(mt - nu)}(stm(u + v) - nuv(s + t))\\ &= \frac{(m - n)}{(ms - nv)(mt - nu)}\left(m \frac{c^2 - r^2}{m^2 + 1} \frac{2cn}{n^2 + 1} - n \frac{c^2 - r^2}{n^2 + 1} \frac{2cm}{m^2 + 1} \right)\\ &= 0, \end{align*}

using the sum and product results from above.

It is possible to interpret this question with a diagram (try to match all the parts of this question with the picture below).

The circle $x^2 + (y-c)^2 = r^2$ appears here in black. Both $r$ and $c$ can be varied. The two lines through the origin $y = mx$ and $y = nx$ are in green.