The equation of the circle which passes throu | vedclass

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Question

(B) The radius of a circle passing through two fixed points is minimum when the line segment joining the two points is the diameter of the circle.Give

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The radius of a circle passing through two fixed points is minimum when the line segment joining the two points is the diameter of the circle.Given points are $A(1, 0)$ and $B(0, 1)$.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Substituting the points $(1, 0)$ and $(0, 1)$:$(x - 1)(x - 0) + (y - 0)(y - 1) = 0$$x(x - 1) + y(y - 1) = 0$$x^2 - x + y^2 - y = 0$$x^2 + y^2 - x - y = 0$Thus,the correct option is $B$.

The equation of the circle which passes through $(1, 0)$ and $(0, 1)$ and has its radius as small as possible,is

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