If the inverse point of the point (-1, 1) wit | vedclass

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(B) The equation of the polar of the point $(-1, 1)$ with respect to the circle $x^2+y^2-2x+2y-1=0$ is given by $x(-1) + y(1) - (x-1) + (y+1) - 1 = 0$

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(B) The equation of the polar of the point $(-1, 1)$ with respect to the circle $x^2+y^2-2x+2y-1=0$ is given by $x(-1) + y(1) - (x-1) + (y+1) - 1 = 0$. This simplifies to $-x + y - x + 1 + y + 1 - 1 = 0$,which is $-2x + 2y + 1 = 0$ or $2x - 2y - 1 = 0$. The inverse point $(p, q)$ is the foot of the perpendicular from the center of the circle to the polar line. The center of the circle $x^2+y^2-2x+2y-1=0$ is $(1, -1)$. The foot of the perpendicular $(p, q)$ from $(x_0, y_0) = (1, -1)$ to the line $ax + by + c = 0$ is given by $\frac{p-x_0}{a} = \frac{q-y_0}{b} = -\frac{ax_0+by_0+c}{a^2+b^2}$. Here $a=2, b=-2, c=-1$. $\frac{p-1}{2} = \frac{q-(-1)}{-2} = -\frac{2(1)-2(-1)-1}{2^2+(-2)^2} = -\frac{2+2-1}{8} = -\frac{3}{8}$. $p-1 = 2(-\frac{3}{8}) = -\frac{3}{4} \Rightarrow p = 1 - \frac{3}{4} = \frac{1}{4}$. $q+1 = -2(-\frac{3}{8}) = \frac{3}{4} \Rightarrow q = \frac{3}{4} - 1 = -\frac{1}{4}$. Thus,$p^2+q^2 = (\frac{1}{4})^2 + (-\frac{1}{4})^2 = \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}$.

If the inverse point of the point $(-1, 1)$ with respect to the circle $x^2+y^2-2x+2y-1=0$ is $(p, q)$,then $p^2+q^2=$

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