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(D) The equation of the circle is $x^2 + y^2 = 8$,so the radius $r = \sqrt{8} = 2\sqrt{2}$.Given point $P = (4, 0)$. The length of the tangent from $P

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$(2, -2)$ or $(-2, 2)$

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(D) The equation of the circle is $x^2 + y^2 = 8$,so the radius $r = \sqrt{8} = 2\sqrt{2}$.Given point $P = (4, 0)$. The length of the tangent from $P$ to the circle is $\sqrt{4^2 + 0^2 - 8} = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2}$.Let the point of tangency be $A(x_1, y_1)$. Since $A$ is on the circle,$x_1^2 + y_1^2 = 8$. The tangent at $A$ is $xx_1 + yy_1 = 8$. Since it passes through $(4, 0)$,$4x_1 = 8 \Rightarrow x_1 = 2$. Substituting into the circle equation,$2^2 + y_1^2 = 8 \Rightarrow y_1^2 = 4$. Since $A$ is in the first quadrant,$y_1 = 2$. Thus,$A = (2, 2)$.We need to find point $B(x, y)$ on the circle such that the chord length $AB = 4$. The distance formula gives $(x - 2)^2 + (y - 2)^2 = 4^2 = 16$.Expanding this: $x^2 - 4x + 4 + y^2 - 4y + 4 = 16$. Since $x^2 + y^2 = 8$,we have $8 - 4x - 4y + 8 = 16$ $\Rightarrow -4x - 4y = 0$ $\Rightarrow y = -x$.Substituting $y = -x$ into $x^2 + y^2 = 8$,we get $x^2 + (-x)^2 = 8$ $\Rightarrow 2x^2 = 8$ $\Rightarrow x^2 = 4$ $\Rightarrow x = \pm 2$.If $x = 2$,then $y = -2$. If $x = -2$,then $y = 2$.Thus,the possible coordinates for $B$ are $(2, -2)$ or $(-2, 2)$.

$A$ tangent drawn from the point $(4, 0)$ to the circle $x^2 + y^2 = 8$ touches it at a point $A$ in the first quadrant. The coordinates of another point $B$ on the circle such that the length of the chord $AB$ is $4$ are:

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