Year: 2015
Paper: 2
Question Number: 7
Course: LFM Pure
Section: Coordinate Geometry
As in previous years the Pure questions were the most popular of the paper with questions 1, 2 and 6 the most popular. The least popular questions on the paper were questions 8, 11 and 13 with fewer than 250 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained given that the answer to be reached had been provided in the question.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
A circle $C$ is said to be \textit{bisected} by a curve $X$ if $X$ meets $C$ in exactly two points and these points are diametrically opposite each other on $C$.
\begin{questionparts}
\item Let $C$ be the circle of radius $a$ in the $x$-$y$ plane with centre at the origin.
Show, by giving its equation, that it is possible to find a circle of given radius $r$ that bisects $C$ provided $r > a$. Show that no circle of radius $r$ bisects $C$ if $r\le a\,$.
\item Let $C_1$ and $C_2$ be circles with centres at $(-d,0)$ and $(d,0)$ and radii $a_1$ and $a_2$, respectively, where $d > a_1$ and $d > a_2$. Let $D$ be a circle of radius $r$ that bisects both $C_1$ and $C_2$. Show that the $x$-coordinate of the centre of $D$ is $\dfrac{a_2^2 - a_1^2}{4d}$.
Obtain an expression in terms of $d$, $r$, $a_1$ and $a_2$ for the $y$-coordinate of the centre of $D$, and deduce that $r$ must satisfy
\[
16r^2d^2 \ge \big (4d^2 +(a_2-a_1)^2\big) \, \big (4d^2 +(a_2+a_1)^2\big)
\,.
\]
\end{questionparts}\begin{questionparts}
\item $C$ has the equation $x^2 + y^2 = a^2$. One suitable circle would ideally pass through $(0,a)$ and $(0,-a)$ have a centre on the positive $x$-axis, so we would need $a^2+c^2 = r^2$ so $c = \sqrt{r^2-a^2}$ and the equation would be $(x-\sqrt{r^2-a^2})^2 + y^2 = r^2$. Clearly a circle with radius $r < a$ cannot pass through two diametrically opposed points of a circle radius $a$, since the furthest two points can be on a circle is $2r$, and diametrically opposed points are $2a$ apart. Similarly if they are exactly the same radii, then if they pass through diametrically opposed points they must be the same circle.
\item Let the centre of $D$ be at $(x,y)$, then it must be a distance of $\sqrt{r^2-a_i}$ from each circle centre, ie
\begin{align*}
&& (x-d)^2+y^2 &= r^2-a_2^2 \\
&& (x+d)^2 + y^2 &= r^2-a_1^2 \\
\Rightarrow && 4dx &= a_2^2 - a_1^2 \\
\Rightarrow && x &= \frac{a_2^2-a_1^2}{4d} \\
\Rightarrow && y^2 &= r^2-a_1^2 - \left (\frac{a_2^2-a_1^2}{4d}+d \right)^2 \\
&&&= r^2 - a_1^2 - \frac{(a_2^2-a_1^2+4d^2)^2}{16d^2} \\
&&&= \frac{16d^2r^2-16d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2+8a_1^2d^2-8a_2^2d^2}{16d^2} \\
&&&= \frac{16d^2r^2-8d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2-8a_2^2d^2}{16d^2} \\
\Rightarrow && y &= \pm \sqrt{ \frac{16d^2r^2-8d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2-8a_2^2d^2}{16d^2}}
\end{align*}
and we need
\begin{align*}
&& 0 &\leq 16d^2r^2-8d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2-8a_2^2d^2 \\
\Rightarrow && 16d^2 d^2 &\geq 8d^2a_1^2 + a_2^4+a_1^4+16d^4+2a_1^2a_2^2+8a_2^2d^2 \\
&&&= (4d^2+(a_2-a_1)^2)(4d^2+(a_2+a_1)^2)
\end{align*}
\end{questionparts}
The first part of this question was generally well answered, although a significant number of answers did not give the equation of the new circle. The case in part (i) where the two circles have the same radius was often not considered and the explanations for there not being such a circle in some cases were often not sufficiently clear. A significant number of candidates made the incorrect assumption in the second part that the centres of the three circles must lie on a straight line or attempted this part of the question with incorrect methods, such as equating the equations of the two given circles. In the final part of the question not all candidates realised that y² must be positive and were unable to obtain the required inequality by any other means.