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

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(C) The radical axis of the circles $x^2+y^2+2 \alpha x+2 \beta y+c=0$ and $x^2+y^2+\frac{3}{2} x+4 y+c=0$ is obtained by subtracting the equations: $

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$4$

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(C) The radical axis of the circles $x^2+y^2+2 \alpha x+2 \beta y+c=0$ and $x^2+y^2+\frac{3}{2} x+4 y+c=0$ is obtained by subtracting the equations: $(2 \alpha - \frac{3}{2})x + (2 \beta - 4)y = 0$ $(4 \alpha - 3)x + 4(\beta - 2)y = 0$ $(4 \alpha - 3)x + (4 \beta - 8)y = 0$. This line touches the circle $x^2+y^2+2x+2y+1=0$,which has center $(-1, -1)$ and radius $r = \sqrt{1^2+1^2-1} = 1$. The perpendicular distance from the center $(-1, -1)$ to the line $(4 \alpha - 3)x + (4 \beta - 8)y = 0$ must equal the radius $1$: $1 = \frac{|(4 \alpha - 3)(-1) + (4 \beta - 8)(-1)|}{\sqrt{(4 \alpha - 3)^2 + (4 \beta - 8)^2}}$ $|-(4 \alpha - 3 + 4 \beta - 8)| = \sqrt{(4 \alpha - 3)^2 + (4 \beta - 8)^2}$ $|-(4 \alpha + 4 \beta - 11)| = \sqrt{(4 \alpha - 3)^2 + (4 \beta - 8)^2}$ Squaring both sides: $(4 \alpha + 4 \beta - 11)^2 = (4 \alpha - 3)^2 + (4 \beta - 8)^2$ $16 \alpha^2 + 16 \beta^2 + 121 + 32 \alpha \beta - 88 \alpha - 88 \beta = 16 \alpha^2 - 24 \alpha + 9 + 16 \beta^2 - 64 \beta + 64$ $32 \alpha \beta - 88 \alpha + 24 \alpha - 88 \beta + 64 \beta = 9 + 64 - 121$ $32 \alpha \beta - 64 \alpha - 24 \beta = -48$ Dividing by $8$: $4 \alpha \beta - 8 \alpha - 3 \beta = -6$ Adding $10$ to both sides: $4 \alpha \beta - 8 \alpha - 3 \beta + 10 = -6 + 10 = 4$.

If the radical axis of the circles $x^2+y^2+2 \alpha x+2 \beta y+c=0$ and $x^2+y^2+\frac{3}{2} x+4 y+c=0$ touches the circle $x^2+y^2+2 x+2 y+1=0$,then $4 \alpha \beta-8 \alpha-3 \beta+10=$

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