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

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(A) The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$. First,normalize the second circle: $x^2+y^2+\frac{3}{2}x+4y+c=0$. Sub

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

Vedclass · Answer

(A) The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$. First,normalize the second circle: $x^2+y^2+\frac{3}{2}x+4y+c=0$. Subtracting the two equations: $(x^2+y^2+2gx+2fy+c) - (x^2+y^2+\frac{3}{2}x+4y+c) = 0$. This simplifies to $x(2g-\frac{3}{2}) + y(2f-4) = 0$. The circle $x^2+y^2+2x+2y+1=0$ can be written as $(x+1)^2+(y+1)^2=1$,so its center is $(-1, -1)$ and radius is $1$. The radical axis touches this circle,so the perpendicular distance from the center $(-1, -1)$ to the line $x(2g-\frac{3}{2}) + y(2f-4) = 0$ must equal 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 $B=0$. $2g-\frac{3}{2}=0 \Rightarrow g=\frac{3}{4}$ or $2f-4=0 \Rightarrow f=2$.

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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