The equation of the radical axis of the circl | vedclass

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(A) Given circles are:${S_1} = {x^2} + {y^2} - 3x - 4y + 5 = 0$ .....$(i)$For the second circle,divide by $2$ to make the coefficients of ${x^2}$ and

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$2x + 2y - 1 = 0$

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(A) Given circles are:${S_1} = {x^2} + {y^2} - 3x - 4y + 5 = 0$ .....$(i)$For the second circle,divide by $2$ to make the coefficients of ${x^2}$ and ${y^2}$ equal to $1$:${S_2} = {x^2} + {y^2} - 5x - 6y + 6 = 0$ .....$(ii)$The equation of the radical axis is given by ${S_1} - {S_2} = 0$.Subtracting $(ii)$ from $(i)$:$({x^2} + {y^2} - 3x - 4y + 5) - ({x^2} + {y^2} - 5x - 6y + 6) = 0$$(-3x + 5x) + (-4y + 6y) + (5 - 6) = 0$$2x + 2y - 1 = 0$.

The equation of the radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$ and $2{x^2} + 2{y^2} - 10x - 12y + 12 = 0$ is:

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