If the points of intersection of the ellipses $x^{2} + 2y^{2} - 6x - 12y + 23 = 0$ and $4x + 2y^{2}

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

Accepted Answer

By family of curve equation of circle will be

$${\Rightarrow S_{1} + \lambda S_{2} = 0 }{\Rightarrow \left( x^{2} + 2y^{2} - 6x - 12y + 23 \right) }{+ \lambda\left( 4x^{2} + 2y^{2} - 20x - 12y + 35 \right) = 0 }$$⇒ for circle coeff of $x^{2} =$ coeff. of $y^{2}$

$$\Rightarrow \lambda = \frac{1}{2} $$So equation of circle is

$$\Rightarrow x^{2} + y^{2} - \frac{16}{3}x - 6y + \frac{27}{2} = 0 $$Centre $\left( \frac{8}{3},3 \right):$ Radius $r = \sqrt{\frac{47}{18}} = r$

$$\therefore ab + 18r^{2} = 8 + 47 = 55$$

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