The polar of the point (5, -1/2) with respect | vedclass

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(B) The equation of the circle is $(x - 2)^2 + y^2 = 4$,which expands to $x^2 - 4x + 4 + y^2 = 4$,or $x^2 + y^2 - 4x = 0$.Comparing this with the gene

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$6x - y - 20 = 0$

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(B) The equation of the circle is $(x - 2)^2 + y^2 = 4$,which expands to $x^2 - 4x + 4 + y^2 = 4$,or $x^2 + y^2 - 4x = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = 0$,and $c = 0$.The polar of a point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$.Substituting $(x_1, y_1) = (5, -1/2)$,$g = -2$,$f = 0$,and $c = 0$:$5x - \frac{1}{2}y - 2(x + 5) + 0(y - 1/2) + 0 = 0$$5x - \frac{y}{2} - 2x - 10 = 0$$3x - \frac{y}{2} - 10 = 0$Multiplying by $2$,we get $6x - y - 20 = 0$.

The polar of the point $(5, -1/2)$ with respect to the circle $(x - 2)^2 + y^2 = 4$ is

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