The power of a point (2, -1) with respect to | vedclass

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(C) The power of a point $(x_1, y_1)$ with respect to a circle $(x-h)^2 + (y-k)^2 = r^2$ is given by $(x_1-h)^2 + (y_1-k)^2 - r^2 = 9$. Given the poin

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$4$

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(C) The power of a point $(x_1, y_1)$ with respect to a circle $(x-h)^2 + (y-k)^2 = r^2$ is given by $(x_1-h)^2 + (y_1-k)^2 - r^2 = 9$. Given the point $(2, -1)$,radius $r=4$,and centre $(\alpha, \beta)$ on the line $x+y=0$,we have $\beta = -\alpha$. Since the centre is in the $2^{	ext{nd}}$ quadrant,$\alpha  0$. Substituting the values: $(2-\alpha)^2 + (-1-\beta)^2 - 4^2 = 9$. Since $\beta = -\alpha$,we have $(2-\alpha)^2 + (-1+\alpha)^2 - 16 = 9$. Expanding: $(4 - 4\alpha + \alpha^2) + (1 - 2\alpha + \alpha^2) - 16 = 9$. $2\alpha^2 - 6\alpha + 5 - 16 = 9 \implies 2\alpha^2 - 6\alpha - 20 = 0$. Dividing by $2$: $\alpha^2 - 3\alpha - 10 = 0$. Factoring: $(\alpha - 5)(\alpha + 2) = 0$. Since $\alpha Then $\beta = -(-2) = 2$. Thus,$\beta - \alpha = 2 - (-2) = 4$.

The power of a point $(2, -1)$ with respect to a circle $C$ of radius $4$ is $9$. The centre of the circle $C$ lies on the line $x+y=0$ and in the $2^{\text{nd}}$ quadrant. If $(\alpha, \beta)$ is the centre of the circle $C$,then $\beta-\alpha=$

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