If a circle of constant radius 3k passes thro | vedclass

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(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy = 0$. Since it passes through the origin,the constant term is $0$. The radius of the circ

Vedclass · Accepted Answer

$x^2 + y^2 = 4k^2$

Vedclass · Answer

(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy = 0$. Since it passes through the origin,the constant term is $0$. The radius of the circle is given by $r = \sqrt{g^2 + f^2} = 3k$. Squaring both sides,we get $g^2 + f^2 = 9k^2$. The circle meets the $x$-axis at $A$ and the $y$-axis at $B$. Setting $y=0$ in the circle equation,$x^2 + 2gx = 0$,so $x = 0$ or $x = -2g$. Thus,$A = (-2g, 0)$. Setting $x=0$ in the circle equation,$y^2 + 2fy = 0$,so $y = 0$ or $y = -2f$. Thus,$B = (0, -2f)$. The centroid $(x, y)$ of $\Delta OAB$ is given by $x = \frac{0 + (-2g) + 0}{3} = -\frac{2g}{3}$ and $y = \frac{0 + 0 + (-2f)}{3} = -\frac{2f}{3}$. Therefore,$g = -\frac{3x}{2}$ and $f = -\frac{3y}{2}$. Substituting these into $g^2 + f^2 = 9k^2$,we get $\left(-\frac{3x}{2}\right)^2 + \left(-\frac{3y}{2}\right)^2 = 9k^2$. $\frac{9x^2}{4} + \frac{9y^2}{4} = 9k^2$. Dividing by $9$,we get $\frac{x^2}{4} + \frac{y^2}{4} = k^2$,which simplifies to $x^2 + y^2 = 4k^2$.

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

If a circle of constant radius $3k$ passes through the origin and meets the axes at $A$ and $B$,the locus of the centroid of the triangle $OAB$ is the circle

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