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Question

(C) The equation of the circle passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda S_2 = 0$. $x^2 - 2x + y^2 - 4y - 4 + \la

Vedclass · Accepted Answer

$-26$

Vedclass · Answer

(C) The equation of the circle passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda S_2 = 0$. $x^2 - 2x + y^2 - 4y - 4 + \lambda(x^2 + 2x + y^2 + 4y - 4) = 0$. $(1 + \lambda)x^2 + (1 + \lambda)y^2 + 2(\lambda - 1)x + 4(\lambda - 1)y - 4(1 + \lambda) = 0$. Dividing by $(1 + \lambda)$,we get $x^2 + y^2 + \frac{2(\lambda - 1)}{1 + \lambda}x + \frac{4(\lambda - 1)}{1 + \lambda}y - 4 = 0$. Since it passes through $(3, 3)$,substitute $x = 3, y = 3$: $9 + 9 + \frac{6(\lambda - 1)}{1 + \lambda} + \frac{12(\lambda - 1)}{1 + \lambda} - 4 = 0$. $14 + \frac{18(\lambda - 1)}{1 + \lambda} = 0 \Rightarrow 14(1 + \lambda) + 18(\lambda - 1) = 0$. $14 + 14\lambda + 18\lambda - 18 = 0$ $\Rightarrow 32\lambda = 4$ $\Rightarrow \lambda = \frac{1}{8}$. Now,$\alpha = \frac{2(\frac{1}{8} - 1)}{1 + \frac{1}{8}} = \frac{2(-\frac{7}{8})}{\frac{9}{8}} = -\frac{14}{9}$. $\beta = \frac{4(\frac{1}{8} - 1)}{1 + \frac{1}{8}} = \frac{4(-\frac{7}{8})}{\frac{9}{8}} = -\frac{28}{9}$. $\gamma = -4$. $3(\alpha + \beta + \gamma) = 3(-\frac{14}{9} - \frac{28}{9} - 4) = 3(-\frac{42}{9} - 4) = 3(-\frac{14}{3} - 4) = -14 - 12 = -26$.

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