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The points of intersection of two ellipses $x^{2}+2 y^{2}-6 x-12 y+20=0$ and $2 x^{2}+y^{2}-10 x-6 y+15=0$ lie on a circle. The centre of the circle is
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Verified Answer
The correct answer is:
$\left(\frac{8}{3}, 3\right)$
$x^{2}+2 y^{2}-6 x-12 y+20+\lambda\left(2 x^{2}+y^{2}-10 x-6 y+15\right)=0$
$\begin{array}{l}
\Rightarrow(1+2 \lambda) \mathrm{x}^{2}+(2+\lambda) \mathrm{y}^{2}-(10 \lambda+6) \mathrm{x}-(6 \lambda+12) \mathrm{y}+15 \lambda+20=0 \\
2+\lambda=1+2 \lambda \\
\Rightarrow \lambda=1
\end{array}$
$\therefore$ Centre $:\left(\frac{8}{3}, 3\right)$
$\begin{array}{l}
\Rightarrow(1+2 \lambda) \mathrm{x}^{2}+(2+\lambda) \mathrm{y}^{2}-(10 \lambda+6) \mathrm{x}-(6 \lambda+12) \mathrm{y}+15 \lambda+20=0 \\
2+\lambda=1+2 \lambda \\
\Rightarrow \lambda=1
\end{array}$
$\therefore$ Centre $:\left(\frac{8}{3}, 3\right)$
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