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(A) The equation of the family of circles passing through the intersection of $S_1: x^2+y^2-2px=0$ and $S_2: x^2+y^2-2qy=0$ is given by $S_1 + \lambda

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$x^2+y^2-3px+qy=0$

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(A) The equation of the family of circles passing through the intersection of $S_1: x^2+y^2-2px=0$ and $S_2: x^2+y^2-2qy=0$ is given by $S_1 + \lambda S_2 = 0$. $(x^2+y^2-2px) + \lambda(x^2+y^2-2qy) = 0$ $(1+\lambda)x^2 + (1+\lambda)y^2 - 2px - 2\lambda qy = 0$ Dividing by $(1+\lambda)$,we get $x^2+y^2 - \frac{2p}{1+\lambda}x - \frac{2\lambda q}{1+\lambda}y = 0$. The centre of this circle is $(\frac{p}{1+\lambda}, \frac{\lambda q}{1+\lambda})$. Since the centre lies on $\frac{x}{p} - \frac{y}{q} = 2$,we substitute the coordinates: $\frac{1}{p} \cdot \frac{p}{1+\lambda} - \frac{1}{q} \cdot \frac{\lambda q}{1+\lambda} = 2$ $\frac{1}{1+\lambda} - \frac{\lambda}{1+\lambda} = 2$ $\frac{1-\lambda}{1+\lambda} = 2 \implies 1-\lambda = 2+2\lambda \implies -3\lambda = 1 \implies \lambda = -\frac{1}{3}$. Substituting $\lambda = -\frac{1}{3}$ into the family equation: $(x^2+y^2-2px) - \frac{1}{3}(x^2+y^2-2qy) = 0$ $3x^2+3y^2-6px - x^2 - y^2 + 2qy = 0$ $2x^2+2y^2-6px+2qy = 0$ $x^2+y^2-3px+qy = 0$.

The equation of the circle passing through the points of intersection of the circles $x^2+y^2-2px=0$ and $x^2+y^2-2qy=0$ and having its centre on the line $\frac{x}{p}-\frac{y}{q}=2$ is:

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