The equation of a circle that passes through | vedclass

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(C) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$.Since the circle passes through the origin $(0,0)$,we have $c=0$.The circle cuts off an

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$x^2+y^2+2x-3y=0$

Vedclass · Answer

(C) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$.Since the circle passes through the origin $(0,0)$,we have $c=0$.The circle cuts off an intercept of $-2$ on the $X$-axis,so it passes through $(-2,0)$. Substituting this into the equation: $(-2)^2 + 0^2 + 2g(-2) + 2f(0) + 0 = 0$ $\Rightarrow 4 - 4g = 0$ $\Rightarrow g = 1$.The circle cuts off an intercept of $3$ on the $Y$-axis,so it passes through $(0,3)$. Substituting this into the equation: $0^2 + 3^2 + 2g(0) + 2f(3) + 0 = 0$ $\Rightarrow 9 + 6f = 0$ $\Rightarrow f = -\frac{3}{2}$.Substituting $g=1$ and $f=-\frac{3}{2}$ into the general equation: $x^2+y^2+2(1)x+2(-\frac{3}{2})y+0=0$.This simplifies to $x^2+y^2+2x-3y=0$.

The equation of a circle that passes through the origin and cuts off intercepts $-2$ and $3$ on the $X$-axis and $Y$-axis respectively is

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