Question

For hyperbola $\frac{(x-1{)}^2}{9}-\frac{(y-1{)}^2}{4}=1$, and circle $x^2+y^2=25$,the concyclic points are$(x_i,y_i)$ $i=1,2,3,4$ then $2x_1+2x_2+2x_3+2x_4=$

• A. $\frac{2}{13}$
• B. $\frac{4}{13}$
• C. $0$
• D. $8$

Answer 1

The correct option is C $0$

For hyperbola, $\frac{{(x-1)}^2}{9}-\frac{{(y-1)}^2}{4}=1$

and for circle $x^2+y^2=25$

The conic points are $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$

A line passes through the centre of the circle. Hence, it passes through diameter of circle and mid-point of the centre of circle.

Centre of hyperbola $(h,k)=(1,1)$ and radius $=6$

$4(x-1{)}^2-9(y-1{)}^2=36$

Centre of circle $(g,f)=\left(0.0\right)$ and radius $=5$

$(x-0{)}^2+(y-0{)}^2=25$

then, $\frac{x_1+x_2+x_3+x_4}{4}=\frac{h+g}{2},\frac{x_1+x_2+x_3+x_4}{4}=\frac{k+f}{2}\Rightarrow \frac{x_1+x_2+x_3+x_4}{2}=1+0\left(1\right),\frac{x_1+x_2+x_3+x_4}{2}=-1\left(2\right)$

$\Rightarrow 2x_1+2x_2+2x_3+2x_4=0$ [By adding $\left(1\right)$ and $\left(2\right)$]