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  • Two points P and Q are taken on the line joining the points A(0,0) and B(3 a, 0) such that A P=P Q=Q B. Circles are drawn on A P, P Q and Q B as diameters. The locus o

Two points P and Q are taken on the line joining the points A(0,0) and B(3 a, 0) such that A P=P Q=Q B. Circles are drawn on A P, P Q and Q B 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

Option: 1

x^2+y^2-3 a x+2 a^2-b^2=0


Option: 2

3\left(x^2+y^2\right)-9 a x+8 a^2-b^2=0


Option: 3

x^2+y^2-5 a x+6 a^2-b^2=0


Option: 4

x^2+y^2-a x-b^2=0


Answers (1)

best_answer

Since AP=PQ=QB. The coordinates of P are (a,0)

and of Q are (2 a, 0) the centre of the circles on AP, PQ and QB as diameters are respectively C_1\left(\frac{a}{2}, 0\right), C_2\left(\frac{3 a}{2}, 0\right) and \: C_3\left(\frac{5 a}{2}, 0\right) and the radius of each one of them is \left(\frac{a}{2}\right)

Hence, the equations of the circles with centre C_1, C_2 \text { and } C_3 are respectively.

\begin{aligned} & \left(x-\frac{a}{2}\right)^2+y^2=\frac{a^2}{4} ;\left(x-\frac{3 a}{2}\right)^2+y^2=\frac{a^2}{4} \\ \\& \text { and }\left(x-\frac{5 a}{2}\right)^2+y^2=\frac{a^2}{4} \end{aligned}

So that, if S(h,k) be any point on the locus, then 

\begin{aligned} & \left(h-\frac{a}{2}\right)^2+\left(h-\frac{3 a}{2}\right)^2+\left(h-\frac{5 a}{2}\right)^2+3\left(k^2-\frac{a^2}{4}\right)=b^2 \\ \\& \Rightarrow \quad 3\left(h^2+k^2\right)-9 a h+8 a^2=b^2 \end{aligned}

Hence, the locus of S(h,k) is 

3\left(x^2+y^2\right)-9 a x+8 a^2-b^2=0 

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shivangi.shekhar

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