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

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

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

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(D) The equation of the first circle is $x^2+y^2+2gx+2fy+c=0$. Dividing the second circle equation $2x^2+2y^2+3x+8y+2c=0$ 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: $(2g-\frac{3}{2})x + (2f-4)y = 0$. This line passes through the origin $(0,0)$. The circle $x^2+y^2+2x+2y+1=0$ can be written as $(x+1)^2+(y+1)^2=1$,which has center $(-1,-1)$ and radius $r=1$. For the line $Ax+By=0$ to touch this circle,the perpendicular distance from the center $(-1,-1)$ to the line must equal the radius $1$. $\frac{|(2g-\frac{3}{2})(-1) + (2f-4)(-1)|}{\sqrt{(2g-\frac{3}{2})^2 + (2f-4)^2}} = 1$. 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 simplifies to $A^2+B^2+2AB = A^2+B^2$. Thus,$2AB = 0$,implying $A=0$ or $B=0$. If $A=0$,$2g-\frac{3}{2}=0 \Rightarrow g=\frac{3}{4}$. If $B=0$,$2f-4=0 \Rightarrow f=2$.

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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