Question

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

• A. Lie on a circle centered at and of radius
• B. Lie on a circle centered at and of radius
• C. Lie on a circle centered at (8,9) and of radius
• D. Are not concyclic

Answer 1

The correct option is A Lie on a circle centered at and of radius

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 points 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 $\therefore \lambda =2$
This gives the equation of the circle as
$6(x^2+y^2)-32x-36y+81=0$ {using (1)}
$\Rightarrow x^2+y^2-\frac{16}{3}x-6y+\frac{27}{2}=0ItscentreisC(\frac{8}{3},3)andradiusisr=\sqrt{\frac{64}{9}+9-\frac{27}{2}}=\frac{1}{3}\sqrt{\frac{47}{2}}$