Question

The points of intersection of the two ellipse $x^2+2y^2-6x-12y+23=0$ and $4x^2+2y^2-20x-12y+35=0$

• A. lie on a circle centered at $(\frac{8}{3},3)$ and of radius $\frac{1}{3}\sqrt{\frac{47}{2}}$ unit
• B. lie on a circle centered at $(\frac{-8}{3},-3)$ and of radius $\frac{1}{3}\sqrt{\frac{47}{2}}$ unit
• C. lie on a circle centered at $(8,9)$ and of radius $\frac{1}{3}\sqrt{\frac{47}{2}}$ unit
• D. are not concyclic

Answer 1

The correct option is A lie on a circle centered at $(\frac{8}{3},3)$ and of radius $\frac{1}{3}\sqrt{\frac{47}{2}}$ unit

If $S_1=0$ and $S_2=0$ are the equations Then $\lambda S_1+S_2=0$ is a second degree curve passing through the point of intersection of $S_1=0$ and $S_2=0$
$\Rightarrow (\lambda +4)x^2+2(\lambda +1)y^2-2(3\lambda +10)x-12(\lambda +1)y+(23\lambda +35)=0$
For it to be a circle, choose $\lambda$ such that the coefficients of $x^2$ and $y^2$ are equal.
$\therefore \lambda =2$
So, these points are concyclic and equation of circle is$6(x^2+y^2)-32x-36y+81=0$
$\Rightarrow x^2+y^2-\frac{16}{3}x-6y+\frac{27}{2}=0$
Its centre is $C(\frac{8}{3},3)$ and of radius $r=\sqrt{\frac{64}{9}+9-\frac{27}{2}}=\frac{1}{3}\sqrt{\frac{47}{2}}$ unit