Let S be a circle concentric with the circle | vedclass

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(C) The given circle is $3x^2+3y^2+x+y-1=0$. Dividing by $3$,we get $x^2+y^2+\frac{1}{3}x+\frac{1}{3}y-\frac{1}{3}=0$. The center is $(h, k) = \left(-

Vedclass · Accepted Answer

$\frac{-29}{18}$

Vedclass · Answer

(C) The given circle is $3x^2+3y^2+x+y-1=0$. Dividing by $3$,we get $x^2+y^2+\frac{1}{3}x+\frac{1}{3}y-\frac{1}{3}=0$. The center is $(h, k) = \left(-\frac{1}{6}, -\frac{1}{6}\right)$ and the radius $r$ is given by $r^2 = h^2+k^2-c = \frac{1}{36}+\frac{1}{36}+\frac{1}{3} = \frac{1+1+12}{36} = \frac{14}{36} = \frac{7}{18}$. The length of the tangent $L$ from point $(2,-2)$ to the circle $x^2+y^2+\frac{1}{3}x+\frac{1}{3}y-\frac{1}{3}=0$ is given by $L^2 = x_1^2+y_1^2+\frac{1}{3}x_1+\frac{1}{3}y_1-\frac{1}{3}$. $L^2 = (2)^2+(-2)^2+\frac{1}{3}(2)+\frac{1}{3}(-2)-\frac{1}{3} = 4+4+\frac{2}{3}-\frac{2}{3}-\frac{1}{3} = 8-\frac{1}{3} = \frac{23}{3}$. Since the radius of circle $S$ is $L$,the radius squared is $R^2 = \frac{23}{3}$. The equation of circle $S$ is $(x+\frac{1}{6})^2+(y+\frac{1}{6})^2 = \frac{23}{3}$. The power of the point $(2,1)$ with respect to circle $S$ is $(x_1+\frac{1}{6})^2+(y_1+\frac{1}{6})^2 - R^2$. $= (2+\frac{1}{6})^2+(1+\frac{1}{6})^2 - \frac{23}{3} = (\frac{13}{6})^2+(\frac{7}{6})^2 - \frac{23}{3} = \frac{169}{36}+\frac{49}{36} - \frac{276}{36} = \frac{218-276}{36} = -\frac{58}{36} = -\frac{29}{18}$.

Let $S$ be a circle concentric with the circle $3x^2+3y^2+x+y-1=0$. If the length of the tangent drawn from a point $(2,-2)$ to the given circle is the radius of the circle $S$,then the power of the point $(2,1)$ with respect to the circle $S$ is

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