The equation of the circle whose diameter is | vedclass

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(D) Given circles are $S_1: x^2+y^2-3x+y-10=0$ and $S_2: x^2+y^2-x+2y-20=0$. The equation of the common chord is $S_1 - S_2 = 0$,which is $-2x-y+10=0$

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$x^2+y^2-9x-2y+20=0$

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(D) Given circles are $S_1: x^2+y^2-3x+y-10=0$ and $S_2: x^2+y^2-x+2y-20=0$. The equation of the common chord is $S_1 - S_2 = 0$,which is $-2x-y+10=0$ or $2x+y-10=0$. The family of circles passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda S_2 = 0$: $x^2(1+\lambda) + y^2(1+\lambda) - x(3+\lambda) + y(1+2\lambda) - 10 - 20\lambda = 0$. Dividing by $(1+\lambda)$,the center of this circle is $(\frac{3+\lambda}{2(1+\lambda)}, \frac{-(1+2\lambda)}{2(1+\lambda)})$. Since the common chord is the diameter,the center must lie on $2x+y-10=0$. Substituting the center coordinates: $2(\frac{3+\lambda}{2(1+\lambda)}) - \frac{1+2\lambda}{2(1+\lambda)} - 10 = 0$. Multiplying by $2(1+\lambda)$: $2(3+\lambda) - (1+2\lambda) - 20(1+\lambda) = 0$. $6 + 2\lambda - 1 - 2\lambda - 20 - 20\lambda = 0$ $\Rightarrow -20\lambda - 15 = 0$ $\Rightarrow \lambda = -\frac{3}{4}$. Substituting $\lambda = -\frac{3}{4}$ into the family equation: $(1 - \frac{3}{4})x^2 + (1 - \frac{3}{4})y^2 - (3 - \frac{3}{4})x + (1 - \frac{6}{4})y - 10 + 15 = 0$. $\frac{1}{4}x^2 + \frac{1}{4}y^2 - \frac{9}{4}x - \frac{2}{4}y + 5 = 0$. Multiplying by $4$,we get $x^2+y^2-9x-2y+20=0$.

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2-3x+y-10=0$ and $x^2+y^2-x+2y-20=0$ is

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