The power of a point (2,0) with respect to a | vedclass

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(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+C=0$.Since the circle passes through $(-1,-1)$,we have $(-1)^2+(-1)^2+2g(-1)+2f(-1)+C=0$,which

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$\sqrt{13}$

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(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+C=0$.Since the circle passes through $(-1,-1)$,we have $(-1)^2+(-1)^2+2g(-1)+2f(-1)+C=0$,which simplifies to $2-2g-2f+C=0$,or $2g+2f-C=2$ $... (i)$.The power of the point $(2,0)$ with respect to the circle is given by $x^2+y^2+2gx+2fy+C = -4$. Substituting $(2,0)$,we get $4+0+4g+0+C=-4$,which implies $4g+C=-8$,or $C=-8-4g$ $... (ii)$.The length of the tangent from $(1,1)$ is $2$,so $\sqrt{1^2+1^2+2g(1)+2f(1)+C}=2$. Squaring both sides,$2+2g+2f+C=4$,or $2g+2f+C=2$ $... (iii)$.Substituting $C$ from $(ii)$ into $(iii)$: $2g+2f-8-4g=2$,which simplifies to $-2g+2f=10$,or $f-g=5$ $... (iv)$.Substituting $C$ from $(ii)$ into $(i)$: $2g+2f-(-8-4g)=2$,which simplifies to $6g+2f=-6$,or $3g+f=-3$ $... (v)$.Solving $(iv)$ and $(v)$: From $(iv)$,$f=g+5$. Substituting into $(v)$,$3g+(g+5)=-3$,so $4g=-8$,which gives $g=-2$.Then $f=-2+5=3$. From $(ii)$,$C=-8-4(-2)=0$.The radius $r = \sqrt{g^2+f^2-C} = \sqrt{(-2)^2+3^2-0} = \sqrt{4+9} = \sqrt{13}$.

The power of a point $(2,0)$ with respect to a circle $S$ is $-4$ and the length of the tangent drawn from the point $(1,1)$ to $S$ is $2$. If the circle $S$ passes through the point $(-1,-1)$,then the radius of the circle $S$ is

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