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(D) The equation of any circle passing through the intersection of the circle $S = x^2 + y^2 - 4x - 6y - 21 = 0$ and the line $L = 3x + 4y + 5 = 0$ is

Vedclass · Accepted Answer

$x^2 + y^2 + 2x + 2y - 11 = 0$

Vedclass · Answer

(D) The equation of any circle passing through the intersection of the circle $S = x^2 + y^2 - 4x - 6y - 21 = 0$ and the line $L = 3x + 4y + 5 = 0$ is given by $S + \lambda L = 0$.$(x^2 + y^2 - 4x - 6y - 21) + \lambda(3x + 4y + 5) = 0$Since this circle passes through the point $(1, 2)$,we substitute $x = 1$ and $y = 2$ into the equation:$(1^2 + 2^2 - 4(1) - 6(2) - 21) + \lambda(3(1) + 4(2) + 5) = 0$$(1 + 4 - 4 - 12 - 21) + \lambda(3 + 8 + 5) = 0$$-32 + 16\lambda = 0$$16\lambda = 32 \Rightarrow \lambda = 2$Substituting $\lambda = 2$ back into the family of circles equation:$(x^2 + y^2 - 4x - 6y - 21) + 2(3x + 4y + 5) = 0$$x^2 + y^2 - 4x - 6y - 21 + 6x + 8y + 10 = 0$$x^2 + y^2 + 2x + 2y - 11 = 0$Thus,the required equation is $x^2 + y^2 + 2x + 2y - 11 = 0$.

The equation of the circle passing through the point $(1, 2)$ and through the points of intersection of $x^2 + y^2 - 4x - 6y - 21 = 0$ and $3x + 4y + 5 = 0$ is given by

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