If P(2,3) and Q(-1,2) are conjugate points wi | vedclass

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(B) Two points $(x_1, y_1)$ and $(x_2, y_2)$ are conjugate with respect to the circle $x^2+y^2+2gx+2fy+c=0$ if $x_1x_2 + y_1y_2 + g(x_1+x_2) + f(y_1+y

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$\frac{3\sqrt{21}}{\sqrt{2}}$

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(B) Two points $(x_1, y_1)$ and $(x_2, y_2)$ are conjugate with respect to the circle $x^2+y^2+2gx+2fy+c=0$ if $x_1x_2 + y_1y_2 + g(x_1+x_2) + f(y_1+y_2) + c = 0$. Given $P(2,3)$ and $Q(-1,2)$ and the circle $x^2+y^2+2gx+3y-2=0$,we have $f = \frac{3}{2}$ and $c = -2$. Substituting the values: $(2)(-1) + (3)(2) + g(2-1) + \frac{3}{2}(3+2) - 2 = 0$. $-2 + 6 + g + \frac{15}{2} - 2 = 0$. $2 + g + 7.5 = 0 \Rightarrow g = -9.5 = -\frac{19}{2}$. The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-\frac{19}{2})^2 + (\frac{3}{2})^2 - (-2)}$. $r = \sqrt{\frac{361}{4} + \frac{9}{4} + 2} = \sqrt{\frac{370}{4} + \frac{8}{4}} = \sqrt{\frac{378}{4}} = \sqrt{\frac{189}{2}} = \sqrt{\frac{9 	imes 21}{2}} = \frac{3\sqrt{21}}{\sqrt{2}}$.

If $P(2,3)$ and $Q(-1,2)$ are conjugate points with respect to the circle $x^2+y^2+2gx+3y-2=0$,then the radius of the circle is

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