The line Ax + By + C = 0 cuts the circle x^2 | vedclass

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(B) The equation of the circle passing through the intersection of the line $Ax + By + C = 0$ and the circle $x^2 + y^2 + ax + by + c = 0$ is given by

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$0$

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(B) The equation of the circle passing through the intersection of the line $Ax + By + C = 0$ and the circle $x^2 + y^2 + ax + by + c = 0$ is given by:$x^2 + y^2 + ax + by + c + \lambda(Ax + By + C) = 0$ ... $(i)$Similarly,the equation of the circle passing through the intersection of the line $A'x + B'y + C' = 0$ and the circle $x^2 + y^2 + a'x + b'y + c' = 0$ is given by:$x^2 + y^2 + a'x + b'y + c' + \mu(A'x + B'y + C') = 0$ ... $(ii)$Since the four points $P, Q, R,$ and $S$ are concyclic,equations $(i)$ and $(ii)$ must represent the same circle. Comparing the coefficients of $x, y,$ and the constant term,we get:$a + \lambda A = a' + \mu A' \implies (a - a') + \lambda A - \mu A' = 0$$b + \lambda B = b' + \mu B' \implies (b - b') + \lambda B - \mu B' = 0$$c + \lambda C = c' + \mu C' \implies (c - c') + \lambda C - \mu C' = 0$This is a system of linear equations in variables $1, \lambda, -\mu$. For a non-trivial solution to exist,the determinant of the coefficients must be zero:$\left| {\begin{array}{*{20}{c}}{a - a'}&A&{A'}\{b - b'}&B&{B'}\{c - c'}&C&{C'}\end{array}} ight| = 0$Transposing the matrix (which does not change the determinant value),we get:$\left| {\begin{array}{*{20}{c}}{a - a'}&{b - b'}&{c - c'}\A&B&C\{A'}&{B'}&{C'}\end{array}} ight| = 0$Thus,$D = 0$.

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