The power of the point B(-1, 1) with respect | vedclass

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(A) The power of the point $B(-1, 1)$ with respect to the circle $S \equiv x^2+y^2-2x-4y+3=0$ is given by substituting the coordinates into the equati

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lies inside the circle $S^{\prime} = 0$

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(A) The power of the point $B(-1, 1)$ with respect to the circle $S \equiv x^2+y^2-2x-4y+3=0$ is given by substituting the coordinates into the equation: $p = (-1)^2 + (1)^2 - 2(-1) - 4(1) + 3 = 1 + 1 + 2 - 4 + 3 = 3$. Since the length of the tangent $t = \sqrt{p}$,we have $t = \sqrt{3}$,so $t^2 = 3$. The circle $S^{\prime}$ has its centre at $(p, t^2) = (3, 3)$ and passes through the origin $(0, 0)$. The radius squared $r^2$ is the distance from $(3, 3)$ to $(0, 0)$: $r^2 = (3-0)^2 + (3-0)^2 = 9 + 9 = 18$. Thus,the equation of circle $S^{\prime}$ is $(x-3)^2 + (y-3)^2 = 18$. To check the position of the point $(2, 3)$ with respect to $S^{\prime}$,we calculate the power of the point: $(2-3)^2 + (3-3)^2 - 18 = (-1)^2 + 0 - 18 = 1 - 18 = -17$. Since the power is negative $(-17 < 0)$,the point $(2, 3)$ lies inside the circle $S^{\prime} = 0$.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

The power of the point $B(-1, 1)$ with respect to the circle $S \equiv x^2+y^2-2x-4y+3=0$ is $p$. If the length of the tangent drawn from $B$ to the circle $S=0$ is $t$,then the point $(2, 3)$ with respect to the circle $S^{\prime}=0$ having centre at $(p, t^2)$ and passing through the origin:

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