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.