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(A) The equation of a family of circles passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$.Gi

Vedclass · Accepted Answer

$23{x^2} + 23{y^2} - 156x + 38y + 168 = 0$

Vedclass · Answer

(A) The equation of a family of circles passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$.Given $S_1: {x^2} + {y^2} - 8x - 2y + 7 = 0$ and $S_2: {x^2} + {y^2} - 4x + 10y + 8 = 0$.The equation is $({x^2} + {y^2} - 8x - 2y + 7) + \lambda ({x^2} + {y^2} - 4x + 10y + 8) = 0$.Since the circle passes through $(3, -3)$,substitute $x = 3$ and $y = -3$ into the equation:$(9 + 9 - 24 + 6 + 7) + \lambda (9 + 9 - 12 - 30 + 8) = 0$$7 + \lambda (-16) = 0$$16\lambda = 7 \Rightarrow \lambda = \frac{7}{16}$.Substituting $\lambda = \frac{7}{16}$ back into the equation:$16({x^2} + {y^2} - 8x - 2y + 7) + 7({x^2} + {y^2} - 4x + 10y + 8) = 0$$16{x^2} + 16{y^2} - 128x - 32y + 112 + 7{x^2} + 7{y^2} - 28x + 70y + 56 = 0$$23{x^2} + 23{y^2} - 156x + 38y + 168 = 0$.

The equation of the circle passing through the point of intersection of the circles ${x^2} + {y^2} - 8x - 2y + 7 = 0$ and ${x^2} + {y^2} - 4x + 10y + 8 = 0$ and the point $(3, -3)$ is:

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