The equation of the circle passing through th | vedclass

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(A) Let the points be $A(1, 0)$ and $B(0, 1)$.For a circle passing through two points to have the smallest radius,the line segment joining these two p

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the points be $A(1, 0)$ and $B(0, 1)$.For a circle passing through two points to have the smallest radius,the line segment joining these two points must be the diameter of the circle.The midpoint of $AB$ is the center of the circle: $C = (\frac{1+0}{2}, \frac{0+1}{2}) = (\frac{1}{2}, \frac{1}{2})$.The diameter is the distance $AB = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1+1} = \sqrt{2}$.The radius $r = \frac{	ext{diameter}}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.The equation of the circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.Substituting the values: $(x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = (\frac{1}{\sqrt{2}})^2$.$x^2 - x + \frac{1}{4} + y^2 - y + \frac{1}{4} = \frac{1}{2}$.$x^2 + y^2 - x - y + \frac{1}{2} = \frac{1}{2}$.$x^2 + y^2 - x - y = 0$.

The equation of the circle passing through the points $(1, 0)$ and $(0, 1)$ with the smallest radius is . . . . . .

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