The equation of the circle which passes throu | vedclass

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Question

(C) Since the circle passes through the origin $(0, 0)$,the constant term $c = 0$.The circle cuts off intercepts of length $2$ on the negative coordin

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Since the circle passes through the origin $(0, 0)$,the constant term $c = 0$.The circle cuts off intercepts of length $2$ on the negative coordinate axes.The length of the intercept on the $x$-axis is $2\sqrt{g^2 - c} = 2$. Since the intercept is on the negative axis,the center's $x$-coordinate $-g$ must be negative,so $g = 1$.The length of the intercept on the $y$-axis is $2\sqrt{f^2 - c} = 2$. Since the intercept is on the negative axis,the center's $y$-coordinate $-f$ must be negative,so $f = 1$.The general equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Substituting $g = 1$,$f = 1$,and $c = 0$,we get $x^2 + y^2 + 2x + 2y = 0$.

The equation of the circle which passes through the origin and cuts off intercepts of $2$ units length from the negative coordinate axes is:

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