Two points $P$ and $Q$ are taken on the line joining the points $A(0,\ 0)$ and $B(3a,\ 0)$ such that$AP = PQ = QB$. Circles are drawn on $AP,\ PQ$ and $QB$ as diameters The locus of the points, the sum of the squares of the tangents from which to the three circles is equal to $b^{2}$, is

Question

Dr. Abubacker · Accepted Answer

(b)
Since,$\ AP = PQ = QB$. The coordinates of $P$ are $(a,\ 0)$ and of $Q$ are $(2a,\ 0)$ the centre of the circles on $AP,\ pQ$ and $QB$ asdiameters are respectively $C_{1}\left( \frac{a}{2},0 \right),\ C_{2}\left( \frac{3a}{2},0 \right)$ and $C_{3}\left( \frac{5a}{2},0 \right)$ and the radius of each one of them is $\left( \frac{a}{2} \right)$
[image:11_image48.emf]
Hence, the equations of the circles with centre $C_{1},C_{2}$ and $C_{3}$ are respectively
$\left( x - \frac{a}{2} \right)^{2} + y^{2} = \frac{a^{2}}{4};\ \left( x - \frac{3a}{2} \right)^{2} + y^{2} = \frac{a^{2}}{4}$
and $\left( x - \frac{5a}{2} \right)^{2} + y^{2} = \frac{a^{2}}{4}$
So that, if $S(h,\ k)$ be any point on the locus, then
$\left( h - \frac{a}{2} \right)^{2} + \left( h - \frac{3a}{2} \right)^{2} + \left( h - \frac{5a}{2} \right)^{2} + 3\left( k^{2} - \frac{a^{2}}{4} \right) = b^{2}$
$\Rightarrow \ 3\left( h^{2} + k^{2} \right) - 9ah + 8a^{2} = b^{2}$
Hence, the locus of $S(h,\ k)$ is
$3\left( x^{2} + y^{2} \right) - 9ax + 8a^{2} - b^{2} = 0$

Q. Two points $P$ and $Q$ are taken on the line joining the points $A(0,\ 0)$ and $B(3a,\ 0)$ such that$AP = PQ = QB$. Circles are drawn on $AP,\ PQ$ and $QB$ as diameters The locus of the points, the sum of the squares of the tangents from which to the three circles is equal to $b^{2}$, is

Explanation

Reviewed by Dr. Abubacker

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