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 \(r\).

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.