The equation of the circle whose diameter is | vedclass

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(C) Let the given circles be $S_1 \equiv x^2+y^2+2x+3y+2=0$ and $S_2 \equiv x^2+y^2+2x-3y-4=0$. The equation of the common chord is given by $S_1 - S_

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$x^2+y^2+2x+2y+1=0$

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(C) Let the given circles be $S_1 \equiv x^2+y^2+2x+3y+2=0$ and $S_2 \equiv x^2+y^2+2x-3y-4=0$. The equation of the common chord is given by $S_1 - S_2 = 0$. $(x^2+y^2+2x+3y+2) - (x^2+y^2+2x-3y-4) = 0$ $6y + 6 = 0 \Rightarrow y = -1$. Substituting $y = -1$ into the equation of $S_1$: $x^2 + (-1)^2 + 2x + 3(-1) + 2 = 0$ $x^2 + 1 + 2x - 3 + 2 = 0$ $x^2 + 2x = 0$ $\Rightarrow x(x+2) = 0$ $\Rightarrow x = 0, -2$. The endpoints of the diameter are $(0, -1)$ and $(-2, -1)$. The equation of the circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$. $(x-0)(x-(-2)) + (y-(-1))(y-(-1)) = 0$ $x(x+2) + (y+1)^2 = 0$ $x^2 + 2x + y^2 + 2y + 1 = 0$ $x^2 + y^2 + 2x + 2y + 1 = 0$.

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2x+3y+2=0$ and $x^2+y^2+2x-3y-4=0$ is

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