The equation of the circle passing through (1 | vedclass

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(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ $(i)$.Since the circle passes through $(1, 0)$ and $(0, 1)$:For $(1, 0)$: $1 + 2

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

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(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ $(i)$.Since the circle passes through $(1, 0)$ and $(0, 1)$:For $(1, 0)$: $1 + 2g + c = 0$ $(ii)$.For $(0, 1)$: $1 + 2f + c = 0$ $(iii)$.Subtracting $(iii)$ from $(ii)$,we get $2g - 2f = 0$,so $g = f$.Substituting $f = g$ into $(ii)$,we get $c = -(1 + 2g)$.The radius $R$ is given by $R = \sqrt{g^2 + f^2 - c} = \sqrt{g^2 + g^2 - (-(1 + 2g))} = \sqrt{2g^2 + 2g + 1}$.To minimize $R$,we minimize $R^2 = 2g^2 + 2g + 1$.Differentiating with respect to $g$: $\frac{d}{dg}(2g^2 + 2g + 1) = 4g + 2 = 0$,which gives $g = -\frac{1}{2}$.Since $g = f$,we have $f = -\frac{1}{2}$.Then $c = -(1 + 2(-\frac{1}{2})) = -(1 - 1) = 0$.Substituting $g, f, c$ into $(i)$,the equation is $x^2 + y^2 - x - y = 0$.Alternatively,the circle with minimum radius passing through two points has the segment joining those points as its diameter.The equation is $(x - 1)(x - 0) + (y - 0)(y - 1) = 0$,which simplifies to $x^2 - x + y^2 - y = 0$ or $x^2 + y^2 - x - y = 0$.

The equation of the circle passing through $(1, 0)$ and $(0, 1)$ with the minimum possible radius is:

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