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Question

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle passes through $(1, \sqrt{2})$,$(7, \sqrt{2})$,and $(1, 3)$:$1$

Vedclass · Accepted Answer

$x^2 + y^2 - 8x - (3 + \sqrt{2})y + 7 + 3\sqrt{2} = 0$

Vedclass · Answer

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle passes through $(1, \sqrt{2})$,$(7, \sqrt{2})$,and $(1, 3)$:$1$) For $(1, \sqrt{2})$: $1 + 2 + 2g + 2\sqrt{2}f + c = 0 \implies 2g + 2\sqrt{2}f + c = -3$.$2$) For $(7, \sqrt{2})$: $49 + 2 + 14g + 2\sqrt{2}f + c = 0 \implies 14g + 2\sqrt{2}f + c = -51$.Subtracting $(1)$ from $(2)$: $12g = -48 \implies g = -4$.$3$) For $(1, 3)$: $1 + 9 + 2g + 6f + c = 0 \implies 2g + 6f + c = -10$.Substitute $g = -4$ into $(1)$: $-8 + 2\sqrt{2}f + c = -3 \implies 2\sqrt{2}f + c = 5$.Substitute $g = -4$ into $(3)$: $-8 + 6f + c = -10 \implies 6f + c = -2$.Subtracting these two: $f(6 - 2\sqrt{2}) = -7 \implies f = \frac{-7}{6 - 2\sqrt{2}} = \frac{-7(6 + 2\sqrt{2})}{36 - 8} = \frac{-7(6 + 2\sqrt{2})}{28} = -\frac{3 + \sqrt{2}}{2}$.Now,$c = -2 - 6f = -2 - 6(-\frac{3 + \sqrt{2}}{2}) = -2 + 3(3 + \sqrt{2}) = -2 + 9 + 3\sqrt{2} = 7 + 3\sqrt{2}$.Substituting $g, f, c$ into the general equation: $x^2 + y^2 - 8x - (3 + \sqrt{2})y + 7 + 3\sqrt{2} = 0$.

The equation of the circle passing through the points $(1, \sqrt{2})$,$(7, \sqrt{2})$,and $(1, 3)$ is:

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