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A circle with center (3 \alpha, 3 \beta) and of variable radius cuts the rectangular hyperbola \mathrm{x^2-y^2=9 a^2} at the points P, Q, R, S. The locus of the centroid of the triangle PQR is

Option: 1

(x+\alpha)^2-(y+\beta)^2=a^2


Option: 2

(x-\alpha)^2-(y-\beta)^2=2 a^2


Option: 3

(x-2 \alpha)^2+(y-2 \beta)^2=a^2


Option: 4

 none of these 


Answers (1)

best_answer

The equation of the circle with center (3 \alpha, 3 \beta) is
\mathrm{x^2+y^2-6 \alpha x-6 \beta y+k=0 }...................(1)
where \mathrm{\mathrm{k} } variable.
Given hyperbola is 
\mathrm{x^2-y^2=9 a^2 }..............................................(2)
\mathrm{\begin{aligned} & \left(\mathrm{x}^2+\mathrm{x}^2-9 \mathrm{a}^2-6 \alpha \mathrm{x}+\mathrm{k}\right)^2=36 \beta^2\left(\mathrm{x}^2-9 \mathrm{a}^2\right) \\ \Rightarrow \quad & 4 \mathrm{x}^4-24 \alpha \mathrm{x}^3+\ldots=0 \end{aligned} }
this is a biquadratic equation.
Let the abscissas of four points P, Q, R & S are  \mathrm{\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3 \& \mathrm{x}_4 } then
\mathrm{x_1+x_2+x_3+x_4=6 \alpha }...............................(3)
Similarly \mathrm{y_1+y_2+y_3+y_4=6 \beta }............................(4)
If \mathrm{\left(x^{\prime}, y^{\prime}\right) } be the centroid of the triangle P Q R, then
\mathrm{\begin{array}{ll} & x^{\prime}=\frac{x_1+x_2+x_3}{3}=\frac{6 \alpha-x_4}{3} \\ \therefore & -3\left(x^{\prime}-2 \alpha\right)-x_4 \\ \text { and } & y^{\prime}=\frac{y_1+y_2+y_3}{3}=\frac{6 \beta-y_4}{3} \\ \therefore & -3\left(y^{\prime}-2 \beta\right)=y_4 \end{array} }
But \mathrm{\quad x_4^2-y_4^2=9 a^2 }
\mathrm{\begin{aligned} & \Rightarrow \quad 9\left(x^{\prime}-2 \alpha\right)^2-9\left(y^{\prime}-2 \beta\right)^2=9 a^2 \\ & \Rightarrow \quad\left(x^{\prime}-2 \alpha\right)^2-\left(y^{\prime}-2 \beta\right)^2=a^2 \end{aligned} }
Hence locus of controid \mathrm{\left(x^{\prime}, y^{\prime}\right) } is
\mathrm{(x-2 \alpha)^2-(y-2 \beta)^2=a^2 }

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