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(C) The family of circles passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$ (where $\lambda

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$7x^2 + 7y^2 - 10x - 10y - 12 = 0$

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(C) The family of circles passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$ (where $\lambda 
eq -1$).Substituting the given equations: $(x^2 + y^2 - 6x + 2y + 4) + \lambda(x^2 + y^2 + 2x - 4y - 6) = 0$.Rearranging the terms: $(1 + \lambda)x^2 + (1 + \lambda)y^2 + (2\lambda - 6)x + (2 - 4\lambda)y + (4 - 6\lambda) = 0$.Dividing by $(1 + \lambda)$: $x^2 + y^2 + \frac{2\lambda - 6}{1 + \lambda}x + \frac{2 - 4\lambda}{1 + \lambda}y + \frac{4 - 6\lambda}{1 + \lambda} = 0$.The centre of this circle is $(-\frac{\lambda - 3}{1 + \lambda}, -\frac{1 - 2\lambda}{1 + \lambda}) = (\frac{3 - \lambda}{1 + \lambda}, \frac{2\lambda - 1}{1 + \lambda})$.Since the centre lies on $y = x$,we have $\frac{3 - \lambda}{1 + \lambda} = \frac{2\lambda - 1}{1 + \lambda}$.This implies $3 - \lambda = 2\lambda - 1$,so $3\lambda = 4$,which gives $\lambda = \frac{4}{3}$.Substituting $\lambda = \frac{4}{3}$ into the family equation: $(x^2 + y^2 - 6x + 2y + 4) + \frac{4}{3}(x^2 + y^2 + 2x - 4y - 6) = 0$.Multiplying by $3$: $3(x^2 + y^2 - 6x + 2y + 4) + 4(x^2 + y^2 + 2x - 4y - 6) = 0$.$3x^2 + 3y^2 - 18x + 6y + 12 + 4x^2 + 4y^2 + 8x - 16y - 24 = 0$.$7x^2 + 7y^2 - 10x - 10y - 12 = 0$.

If the centre of a circle,which passes through the points of intersection of the circles $x^2 + y^2 - 6x + 2y + 4 = 0$ and $x^2 + y^2 + 2x - 4y - 6 = 0$,lies on the line $y = x$,then the equation of the circle is:

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