If the radical axis of the circles x^2+y^2+2g | vedclass

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(A) The equation of the first circle is $x^2+y^2+2gx+2fy+c=0$. Dividing the second circle $2x^2+2y^2+3x+8y+2c=0$ by $2$,we get $x^2+y^2+\frac{3}{2}x+4

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(A) The equation of the first circle is $x^2+y^2+2gx+2fy+c=0$. Dividing the second circle $2x^2+2y^2+3x+8y+2c=0$ by $2$,we get $x^2+y^2+\frac{3}{2}x+4y+c=0$. The radical axis is obtained by subtracting the two equations: $(2g-\frac{3}{2})x + (2f-4)y = 0$,which simplifies to $(4g-3)x + 2(2f-4)y = 0$,or $(4g-3)x + 4(f-2)y = 0$. This line touches the circle $x^2+y^2+2x+2y+1=0$,which can be written as $(x+1)^2+(y+1)^2=1$. The center of this circle is $(-1, -1)$ and the radius is $1$. The perpendicular distance from the center $(-1, -1)$ to the line $(4g-3)x + 4(f-2)y = 0$ must be equal to the radius $1$: $\frac{|(4g-3)(-1) + 4(f-2)(-1)|}{\sqrt{(4g-3)^2 + (4(f-2))^2}} = 1$. $|-(4g-3) - 4(f-2)| = \sqrt{(4g-3)^2 + 16(f-2)^2}$. Squaring both sides: $(4g-3)^2 + 16(f-2)^2 + 8(4g-3)(f-2) = (4g-3)^2 + 16(f-2)^2$. This simplifies to $8(4g-3)(f-2) = 0$,which implies $(4g-3)(f-2) = 0$.

If the radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$,then $(4g-3)(f-2)=$

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