If P(alpha, beta) is the radical centre of th | vedclass

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(D) First,we normalize the equations of the circles to the form $x^2+y^2+2gx+2fy+c=0$: $S \equiv x^2+y^2+4x+7=0$ $S^{\prime} \equiv x^2+y^2+\frac{3}{2

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$2$

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(D) First,we normalize the equations of the circles to the form $x^2+y^2+2gx+2fy+c=0$: $S \equiv x^2+y^2+4x+7=0$ $S^{\prime} \equiv x^2+y^2+\frac{3}{2}x+\frac{5}{2}y+\frac{9}{2}=0$ $S^{\prime \prime} \equiv x^2+y^2+y=0$ The radical axis of $S$ and $S^{\prime \prime}$ is $S-S^{\prime \prime}=0$,which gives $4x-y+7=0$ (Equation $1$). The radical axis of $S^{\prime \prime}$ and $S^{\prime}$ is $S^{\prime \prime}-S^{\prime}=0$,which gives $x^2+y^2+y - (x^2+y^2+\frac{3}{2}x+\frac{5}{2}y+\frac{9}{2}) = 0$,simplifying to $-\frac{3}{2}x-\frac{3}{2}y-\frac{9}{2}=0$,or $x+y+3=0$ (Equation $2$). Solving the system of equations $4x-y+7=0$ and $x+y+3=0$: Adding the two equations: $5x+10=0 \implies x=-2$. Substituting $x=-2$ into $x+y+3=0$: $-2+y+3=0 \implies y=-1$. Thus,the radical centre $P$ is $(-2, -1)$. The length of the tangent from $P(\alpha, \beta)$ to the circle $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{\alpha^2+\beta^2+2g\alpha+2f\beta+c}$. For $S^{\prime} \equiv x^2+y^2+\frac{3}{2}x+\frac{5}{2}y+\frac{9}{2}=0$,the length is $\sqrt{(-2)^2+(-1)^2+\frac{3}{2}(-2)+\frac{5}{2}(-1)+\frac{9}{2}} = \sqrt{4+1-3-2.5+4.5} = \sqrt{4} = 2$.

List-$I$	List-$II$
$(A)$ Point circles of $S_\alpha=0$	$(i)$ do not exist
$(B)$ Point circles of $S_\beta=0$	(ii) intersecting
$(C)$ The circles in $S_\alpha=0$ are	(iii) non-intersecting
$(D)$ The circles in $S_\beta=0$ are	(iv) $(\pm \sqrt{k}, 0)$
	$(v)$ $(0, \pm \sqrt{k})$

If $P(\alpha, \beta)$ is the radical centre of the circles $S \equiv x^2+y^2+4x+7=0$,$S^{\prime} \equiv 2x^2+2y^2+3x+5y+9=0$ and $S^{\prime \prime} \equiv x^2+y^2+y=0$,then the length of the tangent drawn from $P$ to $S^{\prime}=0$ is

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