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(B) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$ $(i)$.Since the circle passes through $(3,2)$,we have $9+4+6g+4f+c=0$,which impli

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$x^2+y^2+6x-8y-15=0$

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(B) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$ $(i)$.Since the circle passes through $(3,2)$,we have $9+4+6g+4f+c=0$,which implies $6g+4f+c+13=0$ $(ii)$.Since the circle bisects the circumference of $x^2+y^2=15$,the common chord is the diameter of $x^2+y^2=15$. The radical axis is $2gx+2fy+c-15=0$. Since the center $(0,0)$ lies on this line,we get $c-15=0$,so $c=15$ is incorrect; rather,the condition for bisecting the circumference is that the radical axis passes through the center $(0,0)$,so $c-15=0$ is wrong. The radical axis is $2gx+2fy+c+15=0$. Thus $c+15=0$,so $c=-15$ $(iii)$.Since the circle cuts $x^2+y^2+4x+6y+3=0$ orthogonally,we use $2g_1g_2+2f_1f_2=c_1+c_2$. Here $g_1=g, f_1=f, c_1=c$ and $g_2=2, f_2=3, c_2=3$. So $2g(2)+2f(3)=c+3$,which gives $4g+6f=c+3$ $(iv)$.Substituting $c=-15$ in $(iv)$,$4g+6f=-12$,so $2g+3f=-6$.Substituting $c=-15$ in $(ii)$,$6g+4f=2$,so $3g+2f=1$.Solving these,$g=3, f=-4$. Thus the equation is $x^2+y^2+6x-8y-15=0$.

The equation of the circle which passes through the point $(3,2)$,bisects the circumference of the circle $x^2+y^2=15$,and cuts the circle $x^2+y^2+4x+6y+3=0$ orthogonally is

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