If the points of intersection of the ellipses x^2+2y^2-6x-12y+23=0 and 4x^2+2y^2-20x-12y+3…

Question

If the points of intersection of the ellipses $x^2+2y^2-6x-12y+23=0$ and $4x^2+2y^2-20x-12y+35=0$ lie on a circle of radius $r$ and centre $(a,b)$, then the value of $ab+18r^2$ is:

Accepted Answer

Let $S_1\equiv x^2+2y^2-6x-12y+23=0$, $S_2\equiv 4x^2+2y^2-20x-12y+35=0$.

The family through the intersection points is $S_1+\lambda S_2=0$. For this to be a circle, the coefficients of $x^2$ and $y^2$ must match:

$$(1+4\lambda)=(2+2\lambda) \;\Rightarrow\; \lambda=\frac12.$$

$$S_1+\tfrac12 S_2 = 3x^2+3y^2-16x-18y+\frac{81}{2}=0 \;\Rightarrow\; x^2+y^2-\frac{16}{3}x-6y+\frac{27}{2}=0.$$

Centre $(a,b)=\left(\dfrac83,3\right)$, and

$$r^2=\left(\frac83\right)^2+3^2-\frac{27}{2}=\frac{64}{9}+9-\frac{27}{2}=\frac{128+162-243}{18}=\frac{47}{18}.$$

$$ab+18r^2 = \frac83\times3+18\times\frac{47}{18}=8+47=55.$$

Solutions · 1