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Question

(C) Let the equation of the family of circles passing through the intersection of $S_1 = x^2 + y^2 - 8x - 2y + 7 = 0$ and $S_2 = x^2 + y^2 - 4x + 10y

Vedclass · Accepted Answer

$x^2 + y^2 + 22y + 9 = 0$

Vedclass · Answer

(C) Let the equation of the family of circles passing through the intersection of $S_1 = x^2 + y^2 - 8x - 2y + 7 = 0$ and $S_2 = x^2 + y^2 - 4x + 10y + 8 = 0$ be $S_1 + \lambda S_2 = 0$.$(x^2 + y^2 - 8x - 2y + 7) + \lambda(x^2 + y^2 - 4x + 10y + 8) = 0$$(1 + \lambda)x^2 + (1 + \lambda)y^2 - (8 + 4\lambda)x + (10\lambda - 2)y + (7 + 8\lambda) = 0$Dividing by $(1 + \lambda)$:$x^2 + y^2 - \frac{8 + 4\lambda}{1 + \lambda}x + \frac{10\lambda - 2}{1 + \lambda}y + \frac{7 + 8\lambda}{1 + \lambda} = 0$The centre of this circle is $(\frac{4 + 2\lambda}{1 + \lambda}, -\frac{5\lambda - 1}{1 + \lambda})$.Since the centre lies on the $y$-axis,the $x$-coordinate of the centre must be $0$.$\frac{4 + 2\lambda}{1 + \lambda} = 0 \implies 4 + 2\lambda = 0 \implies \lambda = -2$.Substituting $\lambda = -2$ into the family equation:$(x^2 + y^2 - 8x - 2y + 7) - 2(x^2 + y^2 - 4x + 10y + 8) = 0$$x^2 + y^2 - 8x - 2y + 7 - 2x^2 - 2y^2 + 8x - 20y - 16 = 0$$-x^2 - y^2 - 22y - 9 = 0$$x^2 + y^2 + 22y + 9 = 0$.

The equation of the circle which passes through the point of intersection of circles $x^2 + y^2 - 8x - 2y + 7 = 0$ and $x^2 + y^2 - 4x + 10y + 8 = 0$ and has its centre on the $y$-axis is:

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