The equation of a circle passing through the | vedclass

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(C) Since the circle passes through the origin $(0,0)$ and makes intercepts of $3$ on the $x$-axis and $-5$ on the $y$-axis,the points $(3,0)$ and $(0

Vedclass · Accepted Answer

$x^{2}+y^{2}-3x+5y=0$

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(C) Since the circle passes through the origin $(0,0)$ and makes intercepts of $3$ on the $x$-axis and $-5$ on the $y$-axis,the points $(3,0)$ and $(0,-5)$ lie on the circle. Since the angle subtended by the diameter at any point on the circle is $90^{\circ}$,the line segment joining $(3,0)$ and $(0,-5)$ acts as the diameter of the circle because the angle at the origin $(0,0)$ is $90^{\circ}$. 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 the points $(3,0)$ and $(0,-5)$: $(x-3)(x-0) + (y-0)(y-(-5)) = 0$ $x(x-3) + y(y+5) = 0$ $x^{2} - 3x + y^{2} + 5y = 0$ $x^{2} + y^{2} - 3x + 5y = 0$

The equation of a circle passing through the origin and making an $x$-intercept of $3$ and a $y$-intercept of $-5$ is

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