If (p, q) is the centre of the circle which c | vedclass

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(A) Let the equation of the required circle be $x^2+y^2-2px-2qy+C=0$. Since this circle cuts the given circles orthogonally,we use the condition $2g_1

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$9$

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(A) Let the equation of the required circle be $x^2+y^2-2px-2qy+C=0$. Since this circle cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$. For the first circle $x^2+y^2-2x-4y+4=0$: $2(-p)(-1) + 2(-q)(-2) = C+4 \Rightarrow 2p+4q = C+4$ $(i)$. For the second circle $x^2+y^2+2x-4y+1=0$: $2(-p)(1) + 2(-q)(-2) = C+1 \Rightarrow -2p+4q = C+1$ $(ii)$. For the third circle $x^2+y^2-4x-2y-11=0$: $2(-p)(-2) + 2(-q)(-1) = C-11 \Rightarrow 4p+2q = C-11$ $(iii)$. Subtracting $(ii)$ from $(i)$: $(2p+4q) - (-2p+4q) = (C+4) - (C+1)$ $\Rightarrow 4p = 3$ $\Rightarrow p = 3/4$. Substituting $p=3/4$ into $(i)$ and $(iii)$: $(i)$ $\Rightarrow 2(3/4) + 4q = C+4$ $\Rightarrow 3/2 + 4q = C+4$ $\Rightarrow 4q - C = 5/2$. $(iii)$ $\Rightarrow 4(3/4) + 2q = C-11$ $\Rightarrow 3 + 2q = C-11$ $\Rightarrow 2q - C = -14$. Subtracting these: $(4q-C) - (2q-C) = 5/2 - (-14)$ $\Rightarrow 2q = 33/2$ $\Rightarrow q = 33/4$. Thus,$p+q = 3/4 + 33/4 = 36/4 = 9$.

If $(p, q)$ is the centre of the circle which cuts the three circles $x^2+y^2-2x-4y+4=0$,$x^2+y^2+2x-4y+1=0$ and $x^2+y^2-4x-2y-11=0$ orthogonally,then $p+q=$

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