The equation of the circle whose diameter is | vedclass

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(A) Let the two circles be $S_1 \equiv x^2+y^2-6x-7=0$ and $S_2 \equiv x^2+y^2-10x+16=0$. The common chord is given by $S_1 - S_2 = 0$. $(x^2+y^2-6x-7

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$8x^2+8y^2-92x+197=0$

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(A) Let the two circles be $S_1 \equiv x^2+y^2-6x-7=0$ and $S_2 \equiv x^2+y^2-10x+16=0$. The common chord is given by $S_1 - S_2 = 0$. $(x^2+y^2-6x-7) - (x^2+y^2-10x+16) = 0$ $4x - 23 = 0 \Rightarrow x = \frac{23}{4}$. The points of intersection are found by substituting $x = \frac{23}{4}$ into $S_1$: $(\frac{23}{4})^2 + y^2 - 6(\frac{23}{4}) - 7 = 0$ $\frac{529}{16} + y^2 - \frac{138}{4} - 7 = 0$ $y^2 = \frac{138}{4} + 7 - \frac{529}{16} = \frac{552 + 112 - 529}{16} = \frac{135}{16}$. So,$y = \pm \frac{3\sqrt{15}}{4}$. The diameter endpoints are $(\frac{23}{4}, \frac{3\sqrt{15}}{4})$ and $(\frac{23}{4}, -\frac{3\sqrt{15}}{4})$. The equation of the circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$. $(x-\frac{23}{4})(x-\frac{23}{4}) + (y-\frac{3\sqrt{15}}{4})(y+\frac{3\sqrt{15}}{4}) = 0$ $(x-\frac{23}{4})^2 + y^2 - \frac{135}{16} = 0$ $x^2 - \frac{23}{2}x + \frac{529}{16} + y^2 - \frac{135}{16} = 0$ $x^2 + y^2 - \frac{23}{2}x + \frac{394}{16} = 0$ $x^2 + y^2 - \frac{23}{2}x + \frac{197}{8} = 0$ Multiplying by $8$: $8x^2 + 8y^2 - 92x + 197 = 0$.

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2-6x-7=0$ and $x^2+y^2-10x+16=0$ is:

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