Question

If a circle and the rectangular hyperbola $xy=c^2$ meet in the four points whose parameters are $p_1,p_2,p_3$ and $p_4$, then the center of the circle through the points $p_1,p_2$ and $p_3$ is

• A. {$\frac{c}{2}(p_1+p_2+p_3+\frac{1}{p_1.p_2.p_3}),\frac{c}{2}(\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}+p_1.p_2.p_3)$}
• B. {$\frac{c}{2}(p_1+p_2+p_3),\frac{c}{2}(p_1.p_2.p_3)$}
• C. {$\frac{c}{2}\left(\frac{1}{p_1.p_2.p_3}\right),\frac{c}{2}(\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}$)}
• D. None of these

Answer 1

The correct option is A {$\frac{c}{2}(p_1+p_2+p_3+\frac{1}{p_1.p_2.p_3}),\frac{c}{2}(\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}+p_1.p_2.p_3)$}

Let the rectangular hyperbola be $xy=c^2$ and the equation of the circle be $x^2+y^2+2gx+2fy+k=0$. Any point on the hyperbola is $(cp,\frac{c}{p})$.If it lies on the circle, then $c^2{p}^2+\frac{c^2}{p^2}+2gcp+2f\frac{c}{p}+k=0$.

$c^2{p}^4+2gc{p}^3+k{p}^2+2fcp+c^2=0$.

This is fourth-degree equation in $p$, which has four roots.

Hence the circle and the hyperbola intersect in four points.

If $p_1,p_2,p_3,p_4$ are the roots of this equation, then$p_1+p_2+p_3+p_4=\frac{2gc}{c^2}=\frac{2g}{c}$ $c{p}_1+c{p}_2+c{p}_3+c{p}_4=-2g$ $\frac{x_1+x_2+x_3+x_4}{4}=\frac{g}{2}$ ..(i)

Also $\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}+\frac{1}{p_4}=\frac{p_1.p_2.p_3}{p_1.p_2.p_3.p_4}=\frac{2fc}{c^2}=\frac{2f}{c}$

$\frac{c}{p_1}+\frac{c}{p_2}+\frac{c}{p_3}+\frac{c}{p_4}=2f$

$\frac{y_1+y_2+y_3+y_4}{4}=f/2$ ..(ii)

Also, $p_1.p_2.p_3.p_4=\frac{c^2}{c^2}=1$ ..(iii)

Hence from (i),(ii) and (iii) we can clearly see that$-g,-f=\frac{c}{2}(p_1+p_2+p_3+\frac{1}{p_1.p_2.p_3}),\frac{c}{2}(\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}+p_1.p_2.p_3)$Hence, the correct option is (A).