Question

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

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

Answer 1

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

Equation of any curve passing through the intersection of the given ellipse is
$4x^2+2y^2-20x-12y+35+\lambda (x^2+2y^2-6x-12y+23)=0$
which represents a circle is
$4+\lambda =2+2\lambda \Rightarrow \lambda =2$
and the equation of the circle is thus,
$6x^2+6y^2-32x-36y+81=0$
$\Rightarrow$ $x^2+y^2-\left(\frac{16}{3}\right)x-6y+\frac{81}{6}=0$
centre of the circle is $(\frac{8}{3},3)$
and the radius is $\sqrt{{\left(\frac{8}{3}\right)}^2+{\left(3\right)}^2-\frac{81}{6}}$
$=\sqrt{\frac{128+162-243}{18}}=\frac{1}{3}\sqrt{\frac{47}{2}}$