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(C) The given circle passes through the origin,so the constant term must be zero: $p - 3 = 0 \Rightarrow p = 3$.Substituting $p = 3$ into the equation

Vedclass · Accepted Answer

$x^{2} + y^{2} + 18x - 6y = 0$

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(C) The given circle passes through the origin,so the constant term must be zero: $p - 3 = 0 \Rightarrow p = 3$.Substituting $p = 3$ into the equation,we get $x^{2} + y^{2} + 9x - 3y = 0$.The radius $r_1$ of this circle is $\sqrt{(\frac{9}{2})^2 + (-\frac{3}{2})^2} = \sqrt{\frac{81}{4} + \frac{9}{4}} = \sqrt{\frac{90}{4}} = \frac{3\sqrt{10}}{2}$.The required circle passes through the origin,so its equation is of the form $x^{2} + y^{2} + 2gx + 2fy = 0$.Its radius $r_2$ is $\sqrt{g^2 + f^2}$.Given $r_2 = 2r_1$,so $\sqrt{g^2 + f^2} = 2 	imes \frac{3\sqrt{10}}{2} = 3\sqrt{10}$.Thus,$g^2 + f^2 = 90$.Since the circle passes through the origin,the tangent at the origin for the first circle is $9x - 3y = 0$,which implies the slope is $3$. For the new circle,the tangent is $gx + fy = 0$,implying the slope is $-g/f$. Assuming the circles are tangent at the origin,$-g/f = 3 \Rightarrow g = -3f$.Substituting into $g^2 + f^2 = 90$: $(-3f)^2 + f^2 = 90$ $\Rightarrow 10f^2 = 90$ $\Rightarrow f^2 = 9$ $\Rightarrow f = \pm 3$.If $f = -3$,then $g = 9$. The equation is $x^{2} + y^{2} + 18x - 6y = 0$.

If a circle passes through the origin and has a radius twice that of the circle $x^{2} + y^{2} + (2p + 3)x + (3 - 2p)y + p - 3 = 0$,find the equation of the circle.

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