If the points (2,3) and (K,-2) are conjugate | vedclass

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(A) The equation of the circle is $x^2+y^2-2x+4y-2=0$. Given that the points $P(2,3)$ and $Q(K,-2)$ are conjugate with respect to the circle,the polar

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$8$

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(A) The equation of the circle is $x^2+y^2-2x+4y-2=0$. Given that the points $P(2,3)$ and $Q(K,-2)$ are conjugate with respect to the circle,the polar of point $P$ must pass through point $Q$. The equation of 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 the values $x_1=2, y_1=3, g=-1, f=2, c=-2$,we get: $x(2)+y(3)-1(x+2)+2(y+3)-2=0$ $2x+3y-x-2+2y+6-2=0$ $x+5y+2=0$. Since the polar passes through $Q(K,-2)$,we substitute $x=K$ and $y=-2$ into the polar equation: $K+5(-2)+2=0$ $K-10+2=0$ $K-8=0$ $K=8$.

If the points $(2,3)$ and $(K,-2)$ are conjugate with respect to the circle $x^2+y^2-2x+4y-2=0$,then $K=$

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