The equation of the circle passing through (0 | vedclass

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(D) The circle passes through the origin $(0,0)$ and makes intercepts $a$ and $b$ on the $x$ and $y$ axes respectively.Therefore,the circle passes thr

Vedclass · Accepted Answer

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

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(D) The circle passes through the origin $(0,0)$ and makes intercepts $a$ and $b$ on the $x$ and $y$ axes respectively.Therefore,the circle passes through the points $(a,0)$ and $(0,b)$.Since the angle between the coordinate axes is $90^{\circ}$,the line segment joining $(a,0)$ and $(0,b)$ is a diameter of the circle.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Substituting $(x_1, y_1) = (a,0)$ and $(x_2, y_2) = (0,b)$:$(x-a)(x-0) + (y-0)(y-b) = 0$$x(x-a) + y(y-b) = 0$$x^2 - ax + y^2 - by = 0$$x^2 + y^2 - ax - by = 0$

The equation of the circle passing through $(0,0)$ and which makes intercepts $a$ and $b$ on the coordinate axes is

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