A circle \(C\) is said to be bisected by a curve \(X\) if \(X\) meets \(C\) in exactly two points and these points are diametrically opposite each other on \(C\).
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 \(r\le a\,\).
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
\[
16r^2d^2 \ge \big (4d^2 +(a_2-a_1)^2\big) \, \big (4d^2 +(a_2+a_1)^2\big)
\,.
\]
Solution:
\(C\) 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.
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*}