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 $x^2+y^2+2gx+2fy+c=0$. The equation of the second circle is $2x^2+2y^2+3x+8y+2c=0$,which can be written as $x^

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either $g=\frac{3}{4}$ or $f=2$

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(A) The equation of the first circle is $x^2+y^2+2gx+2fy+c=0$. The equation of the second circle is $2x^2+2y^2+3x+8y+2c=0$,which can be written as $x^2+y^2+\frac{3}{2}x+4y+c=0$. The radical axis is obtained by subtracting the two equations: $(2g-\frac{3}{2})x + (2f-4)y = 0$. This line passes through the origin $(0,0)$. The third circle is $x^2+y^2+2x+2y+1=0$,which is $(x+1)^2+(y+1)^2=1$. Its center is $(-1,-1)$ and radius is $1$. The line $(2g-\frac{3}{2})x + (2f-4)y = 0$ touches this circle if the perpendicular distance from $(-1,-1)$ to the line is equal to the radius $1$. $\frac{|-(2g-\frac{3}{2})-(2f-4)|}{\sqrt{(2g-\frac{3}{2})^2+(2f-4)^2}} = 1$. Squaring both sides: $(-2g+\frac{3}{2}-2f+4)^2 = (2g-\frac{3}{2})^2+(2f-4)^2$. Let $A = 2g-\frac{3}{2}$ and $B = 2f-4$. Then $(-A-B)^2 = A^2+B^2$,which implies $A^2+B^2+2AB = A^2+B^2$,so $2AB=0$. Thus,$A=0$ or $B=0$. $2g-\frac{3}{2}=0 \implies g=\frac{3}{4}$ or $2f-4=0 \implies f=2$.

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

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