The equation of the radical axis of the circl | vedclass

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(C) Given circles are $S_1: x^2+y^2+4x+6y+7=0$ and $S_2: 4x^2+4y^2+8x+12y-9=0$. To find the radical axis,we first normalize the equation of $S_2$ by d

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$8x+12y+37=0$

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(C) Given circles are $S_1: x^2+y^2+4x+6y+7=0$ and $S_2: 4x^2+4y^2+8x+12y-9=0$. To find the radical axis,we first normalize the equation of $S_2$ by dividing by $4$: $S_2: x^2+y^2+2x+3y-\frac{9}{4}=0$. The equation of the radical axis is given by $S_1 - S_2 = 0$. $(x^2+y^2+4x+6y+7) - (x^2+y^2+2x+3y-\frac{9}{4}) = 0$. $(4x-2x) + (6y-3y) + (7 + \frac{9}{4}) = 0$. $2x + 3y + \frac{28+9}{4} = 0$. $2x + 3y + \frac{37}{4} = 0$. Multiplying by $4$,we get $8x + 12y + 37 = 0$.

The equation of the radical axis of the circles $x^2+y^2+4x+6y+7=0$ and $4x^2+4y^2+8x+12y-9=0$ is:

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