Review question

# Can we find the circles through the points where two ellipses meet? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8216

## Solution

Show that the equation of any circle passing through the points of intersection of the ellipse $(x + 2)^2 + 2y^2 = 18$ and the ellipse $9(x - 1)^2 + 16y^2 = 25$ can be written in the form $x^2 - 2ax + y^2 = 5 - 4a.$

To get some sense of what is going on in this question, it is helpful to sketch a graph.

Both ellipses are symmetrical in the $x$-axis, since replacing $y$ with $-y$ makes no difference to the equations.

The second ellipse is symmetrical in the line $x=1$, since if we substitute in $x=1+k$ and $x=1-k$ into the equation, we obtain the same values for $y$. Likewise, the first ellipse is symmetrical in $x = -2$. (For more on this idea, have a look at Quadratic symmetry.)

If we put $x = 0$, and then $y = 0$ into each equation, we can find the intercepts with the axes.

Alternatively, we could observe that the centre of the first ellipse lies at $(-2,0)$ and its semi-axes have length $\sqrt{18}=3\sqrt{2}$ in the $x$-direction and $\sqrt{9}=3$ in the $y$-direction. Likewise, the second ellipse has centre $(1,0)$ and semi-axes $\frac53$ and $\frac54$.

Either of these approaches gives us enough information to sketch the ellipses.

(Though it is not obvious without a calculator whether $-2+3\sqrt{2}<\frac83$ or not, the question does tell us that the ellipses intersect. Our calculation of their point of intersection will then prove that they do, indeed, intersect, and it will follow that $-2+3\sqrt{2}<\frac83$.)

We can now tackle the question asked.

We find the points of intersection of the two ellipses by solving the simultaneous equations $(x + 2)^2 + 2y^2 = 18$ and $9(x - 1)^2 + 16y^2 = 25.$

We can eliminate $y$ by multiplying the first equation by $8$ and subtracting the second equation, giving \begin{align*} && 8(x + 2)^2 - 9(x - 1)^2 &= 144 - 25&&\quad\\ \iff\quad && 8x^2 + 32x + 32 - 9x^2 + 18x - 9 &= 119\\ \iff\quad && x^2 - 50x + 96 &= 0\\ \iff\quad && (x - 2)(x - 48) &= 0. \end{align*}

This gives the $x$-value of the points of intersection as either $2$ or $48$.

It is clear from our sketch that $x=48$ does not correspond to real points of intersection. (There are, though, complex points of intersection.) So we must have $x=2$. (We could also find this by substituting $x=48$ into the equation of the ellipse and discovering that there are no real values of $y$ corresponding to it.)

Taking $x = 2$ and substituting into the first simultaneous equation, we have \begin{align*} && 16 + 2y^2 &= 18&&\quad\\ \iff\quad&& y^2 &= 1\\ \iff\quad&& y &= \pm1. \end{align*}

Hence there are two (real) points of intersection, at $(2,1)$ and $(2,-1)$. (It’s easy to check that these points really do lie on both ellipses).

A circle through both these points has centre $(a,0)$ on the $x$-axis, by symmetry.

Then the equation of the circle is $(x - a)^2 + y^2 = (2-a)^2+1.$ The right-hand side here is the distance of the centre from $(2,1)$, or from $(2,-1)$, squared.

Expanding gives \begin{align*} && x^2 - 2ax + a^2 + y^2 &= 5 - 4a + a^2&&\quad\\ \iff\quad&& x^2 - 2ax + y^2 &= 5 - 4a, \end{align*}

as required.

These circles form a system of coaxal circles; you might like to explore this concept further.