The points of intersection of two ellipses x^ | vedclass

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(C) The equation of the family of curves passing through the intersection of the two ellipses is given by $S_1 + \lambda S_2 = 0$, where $S_1: x^{2}+2

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$\left(\frac{8}{3}, 3\right)$

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(C) The equation of the family of curves passing through the intersection of the two ellipses is given by $S_1 + \lambda S_2 = 0$, where $S_1: x^{2}+2y^{2}-6x-12y+20=0$ and $S_2: 2x^{2}+y^{2}-10x-6y+15=0$.$(x^{2}+2y^{2}-6x-12y+20) + \lambda(2x^{2}+y^{2}-10x-6y+15) = 0$$(1+2\lambda)x^{2} + (2+\lambda)y^{2} - (6+10\lambda)x - (12+6\lambda)y + (20+15\lambda) = 0$For this to represent a circle, the coefficients of $x^{2}$ and $y^{2}$ must be equal:$1+2\lambda = 2+\lambda \Rightarrow \lambda = 1$Substituting $\lambda = 1$ into the equation:$3x^{2} + 3y^{2} - 16x - 18y + 35 = 0$$x^{2} + y^{2} - \frac{16}{3}x - 6y + \frac{35}{3} = 0$The centre of the circle is given by $\left(-\frac{g}{2}, -\frac{f}{2}\right)$, where $2g = -\frac{16}{3}$ and $2f = -6$.Centre $= \left(\frac{16/3}{2}, \frac{6}{2}\right) = \left(\frac{8}{3}, 3\right)$.

The points of intersection of two ellipses $x^{2}+2y^{2}-6x-12y+20=0$ and $2x^{2}+y^{2}-10x-6y+15=0$ lie on a circle. The centre of the circle is

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