If (3,-2) is the centre of the circle S equiv | vedclass

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(D) The equation of the circle is $x^2+y^2+2gx+2fy-23=0$. The centre is $(-g, -f) = (3, -2)$,so $g = -3$ and $f = 2$. The equation becomes $x^2+y^2-6x

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$(-\frac{9}{5}, -\frac{28}{5})$

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(D) The equation of the circle is $x^2+y^2+2gx+2fy-23=0$. The centre is $(-g, -f) = (3, -2)$,so $g = -3$ and $f = 2$. The equation becomes $x^2+y^2-6x+4y-23=0$. To find the radius $r$,$r = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+2^2-(-23)} = \sqrt{9+4+23} = \sqrt{36} = 6$. The point $A$ on the circle closest to $P(-1, -5)$ lies on the line segment connecting the centre $C(3, -2)$ and $P(-1, -5)$. The vector $\vec{CP} = (-1-3, -5-(-2)) = (-4, -3)$. The distance $CP = \sqrt{(-4)^2+(-3)^2} = 5$. Since $CP However,for a point $P$ inside,the closest point $A$ is on the line $CP$ extended such that $A$ is on the circumference. $A = C + \frac{r}{CP} \vec{CP} = (3, -2) + \frac{6}{5}(-4, -3) = (3 - \frac{24}{5}, -2 - \frac{18}{5}) = (-\frac{9}{5}, -\frac{28}{5})$.

If $(3,-2)$ is the centre of the circle $S \equiv x^2+y^2+2gx+2fy-23=0$ and $A$ is a point on the circle $S=0$ such that its distance from a point $P(-1,-5)$ is least,then $A=$

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