If the chord of contact of the point P(h, k) | vedclass

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(D) The equation of the chord of contact of the point $P(h, k)$ with respect to the circle $x^2+y^2-4x-4y+8=0$ is given by $xh+yk-2(x+h)-2(y+k)+8=0$.

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$(2, 2)$

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(D) The equation of the chord of contact of the point $P(h, k)$ with respect to the circle $x^2+y^2-4x-4y+8=0$ is given by $xh+yk-2(x+h)-2(y+k)+8=0$. Rearranging the terms,we get $(h-2)x + (k-2)y - 2h - 2k + 8 = 0$. The slope of this line is $m = -\frac{h-2}{k-2}$. Given that the line makes an angle of $45^{\circ}$ with the positive $X$-axis,its slope is $	an 45^{\circ} = 1$. Thus,$-\frac{h-2}{k-2} = 1$,which implies $h-2 = -(k-2)$,or $h+k=4$. For the chord to meet the circle in two distinct points,the distance from the center $(2, 2)$ to the line $(h-2)x + (k-2)y - 2h - 2k + 8 = 0$ must be less than the radius $r = \sqrt{2^2+2^2-8} = 0$. Wait,the radius is $r = \sqrt{4+4-8} = 0$. This implies the circle is a point circle at $(2, 2)$. However,if the chord meets the circle at two distinct points,the radius must be greater than $0$. Re-evaluating the circle equation $x^2+y^2-4x-4y+8=0 \Rightarrow (x-2)^2 + (y-2)^2 = 0$. Since the radius is $0$,the only point on the circle is $(2, 2)$. Thus,the chord of contact cannot meet the circle at two distinct points. Given the options,$(2, 2)$ is the center,and the chord of contact for the center is undefined. Therefore,$(2, 2)$ is the point that does not satisfy the condition.

If the chord of contact of the point $P(h, k)$ with respect to the circle $x^2+y^2-4x-4y+8=0$ meets the circle in two distinct points and it also makes an angle $45^{\circ}$ with the positive $X$-axis in the positive direction,then $(h, k)$ cannot be

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