Let S equiv x^2+y^2-6x-6y+4=0 and S^{prime} e | vedclass

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(B) The radical axis of $S=0$ and $S^{\prime}=0$ is given by $S-S^{\prime}=0$. $ (x^2+y^2-6x-6y+4) - (x^2+y^2-2x-4y+3) = 0 $ $ -4x-2y+1=0 \Rightarrow

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$\left(-\frac{19}{5}, -\frac{2}{5}\right)$

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(B) The radical axis of $S=0$ and $S^{\prime}=0$ is given by $S-S^{\prime}=0$. $ (x^2+y^2-6x-6y+4) - (x^2+y^2-2x-4y+3) = 0 $ $ -4x-2y+1=0 \Rightarrow 4x+2y-1=0 $. Any circle belonging to the coaxial system of $S$ and $S^{\prime}$ is given by $S+\lambda(S-S^{\prime})=0$. $ (x^2+y^2-6x-6y+4) + \lambda(4x+2y-1) = 0 $ $ x^2+y^2 + x(4\lambda-6) + y(2\lambda-6) + (4-\lambda) = 0 $. The centre is $C = (3-2\lambda, 3-\lambda)$. The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(3-2\lambda)^2 + (3-\lambda)^2 - (4-\lambda)} = \sqrt{14}$. $ (9-12\lambda+4\lambda^2) + (9-6\lambda+\lambda^2) - 4 + \lambda = 14 $. $ 5\lambda^2 - 17\lambda + 14 = 14 \Rightarrow 5\lambda^2 - 17\lambda = 0 $. Since $\lambda 
eq 0$ (as $\lambda=0$ gives circle $S$),we have $\lambda = \frac{17}{5}$. The centre is $C = (3-2(\frac{17}{5}), 3-\frac{17}{5}) = (3-\frac{34}{5}, \frac{15-17}{5}) = (-\frac{19}{5}, -\frac{2}{5})$.

Let $S \equiv x^2+y^2-6x-6y+4=0$ and $S^{\prime} \equiv x^2+y^2-2x-4y+3=0$ be two circles. The centre of a circle of radius $\sqrt{14}$ and having the same radical axis as $S=0$ and $S^{\prime}=0$ is

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