The radical centre of the circles x^2 + y^2 + | vedclass

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(A) Let the equations of the circles be:$S_1: x^2 + y^2 + 4x + 6y - 19 = 0$$S_2: x^2 + y^2 - 9 = 0$$S_3: x^2 + y^2 - 2x - 2y - 5 = 0$The radical axis

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$(1, 1)$

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(A) Let the equations of the circles be:$S_1: x^2 + y^2 + 4x + 6y - 19 = 0$$S_2: x^2 + y^2 - 9 = 0$$S_3: x^2 + y^2 - 2x - 2y - 5 = 0$The radical axis of $S_1$ and $S_2$ is $S_1 - S_2 = 0$:$(x^2 + y^2 + 4x + 6y - 19) - (x^2 + y^2 - 9) = 0$$4x + 6y - 10 = 0 \implies 2x + 3y = 5$ ... $(i)$The radical axis of $S_2$ and $S_3$ is $S_2 - S_3 = 0$:$(x^2 + y^2 - 9) - (x^2 + y^2 - 2x - 2y - 5) = 0$$2x + 2y - 4 = 0 \implies x + y = 2$ ... $(ii)$Solving equations $(i)$ and $(ii)$:From $(ii)$,$y = 2 - x$. Substituting in $(i)$:$2x + 3(2 - x) = 5$$2x + 6 - 3x = 5$$-x = -1 \implies x = 1$Substituting $x = 1$ in $(ii)$,$1 + y = 2 \implies y = 1$.Thus,the radical centre is $(1, 1)$.

The radical centre of the circles $x^2 + y^2 + 4x + 6y = 19$,$x^2 + y^2 = 9$,and $x^2 + y^2 - 2x - 2y = 5$ is:

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