If the points of intersection of the ellipses | vedclass

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(D) Let the equations of the ellipses be $S_{1} = x^{2}+2y^{2}-6x-12y+23=0$ and $S_{2} = 4x^{2}+2y^{2}-20x-12y+35=0$. Any curve passing through the in

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$55$

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(D) Let the equations of the ellipses be $S_{1} = x^{2}+2y^{2}-6x-12y+23=0$ and $S_{2} = 4x^{2}+2y^{2}-20x-12y+35=0$. Any curve passing through the intersection of these two ellipses is given by $S_{1} + \lambda S_{2} = 0$. $(x^{2}+2y^{2}-6x-12y+23) + \lambda(4x^{2}+2y^{2}-20x-12y+35) = 0$. $(1+4\lambda)x^{2} + (2+2\lambda)y^{2} - (6+20\lambda)x - (12+12\lambda)y + (23+35\lambda) = 0$. For this to be a circle, the coefficient of $x^{2}$ must equal the coefficient of $y^{2}$: $1+4\lambda = 2+2\lambda$ $\Rightarrow 2\lambda = 1$ $\Rightarrow \lambda = \frac{1}{2}$. Substituting $\lambda = \frac{1}{2}$ into the equation: $(1+2)x^{2} + (2+1)y^{2} - (6+10)x - (12+6)y + (23+17.5) = 0$. $3x^{2} + 3y^{2} - 16x - 18y + 40.5 = 0$. Dividing by $3$: $x^{2} + y^{2} - \frac{16}{3}x - 6y + 13.5 = 0$. The centre $(a, b)$ is $(\frac{8}{3}, 3)$. The radius $r$ is given by $r^{2} = g^{2} + f^{2} - c = (\frac{8}{3})^{2} + 3^{2} - 13.5 = \frac{64}{9} + 9 - \frac{27}{2} = \frac{128 + 162 - 243}{18} = \frac{47}{18}$. Thus, $ab + 18r^{2} = (\frac{8}{3} 	imes 3) + 18(\frac{47}{18}) = 8 + 47 = 55$.

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:

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