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(A) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. Since the circle touches the $y$-axis at $(0, 4)$,the center is $(\pm r, 4)$ and the radi

Vedclass · Accepted Answer

$x^2+y^2 \pm 10x - 8y + 16 = 0$

Vedclass · Answer

(A) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. Since the circle touches the $y$-axis at $(0, 4)$,the center is $(\pm r, 4)$ and the radius $r = 4$. Thus,$g^2 = r^2 = 16$ and $f = \pm 4$. Since the circle touches the $y$-axis at $(0, 4)$,the constant term $c$ must satisfy $c = f^2 = 16$. The $x$-intercept is given by $2\sqrt{g^2-c} = 6$,which implies $\sqrt{g^2-c} = 3$,so $g^2-c = 9$. Substituting $c = 16$,we get $g^2 = 16+9 = 25$,so $g = \pm 5$. Substituting $g = \pm 5$,$f = \pm 4$,and $c = 16$ into the general equation: $x^2+y^2 \pm 10x \pm 8y + 16 = 0$. Given the options,the correct equation is $x^2+y^2 \pm 10x - 8y + 16 = 0$.

The circle touching the $y$-axis at a distance $4$ units from the origin and cutting off an intercept $6$ from the $x$-axis is

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