Consider the system of circles x^2+y^2+2fy+la | vedclass

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(B) The given system of circles is $x^2+y^2+2fy+\lambda(x^2+y^2+2gx+k)=0$. Rearranging the terms,we get $(1+\lambda)x^2+(1+\lambda)y^2+2g\lambda x+2fy

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$\frac{k}{2}$

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(B) The given system of circles is $x^2+y^2+2fy+\lambda(x^2+y^2+2gx+k)=0$. Rearranging the terms,we get $(1+\lambda)x^2+(1+\lambda)y^2+2g\lambda x+2fy+\lambda k=0$. Dividing by $(1+\lambda)$,we have $x^2+y^2+2\left(\frac{g\lambda}{1+\lambda}\right)x+2\left(\frac{f}{1+\lambda}\right)y+\frac{\lambda k}{1+\lambda}=0$. For a point circle,the radius $r=0$,so $g'^2+f'^2-c'=0$. Substituting the values,we get $\left(\frac{g\lambda}{1+\lambda}\right)^2+\left(\frac{f}{1+\lambda}\right)^2-\frac{\lambda k}{1+\lambda}=0$. Multiplying by $(1+\lambda)^2$,we get $g^2\lambda^2+f^2-\lambda k(1+\lambda)=0$,which simplifies to $(g^2-k)\lambda^2-k\lambda+f^2=0$. Let the roots be $\lambda_1$ and $\lambda_2$. The centers of the point circles are $A\left(\frac{-g\lambda_1}{1+\lambda_1}, \frac{-f}{1+\lambda_1}\right)$ and $B\left(\frac{-g\lambda_2}{1+\lambda_2}, \frac{-f}{1+\lambda_2}\right)$. Since $\angle AOB = \frac{\pi}{2}$,the product of slopes $m_1 m_2 = -1$. $m_1 = \frac{-f/(1+\lambda_1)}{-g\lambda_1/(1+\lambda_1)} = \frac{f}{g\lambda_1}$. Thus,$\left(\frac{f}{g\lambda_1}\right)\left(\frac{f}{g\lambda_2}\right) = -1$ $\Rightarrow \frac{f^2}{g^2\lambda_1\lambda_2} = -1$. From the quadratic equation,$\lambda_1\lambda_2 = \frac{f^2}{g^2-k}$. Substituting this,$\frac{f^2}{g^2(f^2/(g^2-k))} = -1 \Rightarrow \frac{g^2-k}{g^2} = -1$. $g^2-k = -g^2$ $\Rightarrow 2g^2 = k$ $\Rightarrow g^2 = \frac{k}{2}$.

Consider the system of circles $x^2+y^2+2fy+\lambda(x^2+y^2+2gx+k)=0$,where $g \neq 0, f \neq 0$ and $\lambda$ is a parameter. If $A$ and $B$ are the point circles of this system such that $\angle AOB = \frac{\pi}{2}$,then $g^2$ is equal to

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