If S = x^2 + y^2 + 2x + 17y + 4 = 0,S' = x^2 | vedclass

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(B) The radical axis of $S$ and $S'$ is given by $S - S' = 0$: $(x^2 + y^2 + 2x + 17y + 4) - (x^2 + y^2 + 7x + 6y + 11) = 0$ $-5x + 11y - 7 = 0 \impli

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$\sqrt{57}$

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(B) The radical axis of $S$ and $S'$ is given by $S - S' = 0$: $(x^2 + y^2 + 2x + 17y + 4) - (x^2 + y^2 + 7x + 6y + 11) = 0$ $-5x + 11y - 7 = 0 \implies 5x - 11y + 7 = 0$ ...$(i)$ The radical axis of $S'$ and $S''$ is given by $S' - S'' = 0$: $(x^2 + y^2 + 7x + 6y + 11) - (x^2 + y^2 - x + 22y + 3) = 0$ $8x - 16y + 8 = 0 \implies x - 2y + 1 = 0$ ...(ii) Solving equations $(i)$ and (ii): From (ii),$x = 2y - 1$. Substituting into $(i)$: $5(2y - 1) - 11y + 7 = 0 10y - 5 - 11y + 7 = 0 -y + 2 = 0 \implies y = 2$. Then $x = 2(2) - 1 = 3$. The radical center is $(3, 2)$. The length of the tangent from $(3, 2)$ to $S = 0$ is $\sqrt{S(3, 2)}$: $\sqrt{3^2 + 2^2 + 2(3) + 17(2) + 4} = \sqrt{9 + 4 + 6 + 34 + 4} = \sqrt{57}$.

If $S = x^2 + y^2 + 2x + 17y + 4 = 0$,$S' = x^2 + y^2 + 7x + 6y + 11 = 0$,and $S'' = x^2 + y^2 - x + 22y + 3 = 0$ are three circles,then the length of the tangent from their radical center to $S = 0$ is ......... units.

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