question_answer

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

A) \[{{x}^{2}}+{{y}^{2}}-3ax+2{{a}^{2}}-{{b}^{2}}=0\]

B) \[3({{x}^{2}}+{{y}^{2}})-9ax+8{{a}^{2}}-{{b}^{2}}=0\]

C)   \[{{x}^{2}}+{{y}^{2}}-5ax+6{{a}^{2}}-{{b}^{2}}=0\]

D) \[{{x}^{2}}+{{y}^{2}}-ax-{{b}^{2}}=0\]

Correct Answer: B

Solution :

Since \[AP=PQ=QB.\] the coordinates of \[P\]are (a, 0) and of \[Q\]are (2a, 0). The Centre?s of circle on\[AP\],\[PQ\] and \[QB\]as diameters 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 \[\left( \frac{a}{2} \right)\].

Hence the equation of the circles with entres \[{{C}_{1}},{{C}_{2}}\] and \[{{C}_{3}}\]are respectively.

\[{{\left( x-\frac{a}{2} \right)}^{2}}+{{y}^{2}}=\frac{{{a}^{2}}}{4};\] \[{{\left( x-\frac{3{{a}^{2}}}{2} \right)}^{2}}+{{y}^{2}}=\frac{{{a}^{2}}}{4}\] and \[{{\left( x-\frac{5a}{2} \right)}^{2}}+{{y}^{2}}=\frac{{{a}^{2}}}{4}\]

So that if \[S\left( h,k \right)\]be any point on the locus then

\[{{\left( h-\frac{a}{2} \right)}^{2}}+{{\left( h-\frac{3a}{2} \right)}^{2}}+{{\left( h-\frac{5a}{2} \right)}^{2}}+3\left( {{k}^{2}}-\frac{{{a}^{2}}}{4} \right)={{b}^{2}}\]

\[\Rightarrow 3({{h}^{2}}+{{k}^{2}})-9ah+8{{a}^{2}}={{b}^{2}}\]

So the locus of the \[S(h,k)\]is

\[3\left( {{x}^{2}}+{{y}^{2}} \right)-9ax+8{{a}^{2}}-{{b}^{2}}=0.\]