If the inverse point of the point P(3, 3) wit | vedclass

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(C) The given equation of the circle is $x^2+y^2-4x+4y+4=0$.Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=2, c=4$.The center $O$ is $(-g,

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-$4$

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(C) The given equation of the circle is $x^2+y^2-4x+4y+4=0$.Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=2, c=4$.The center $O$ is $(-g, -f) = (2, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+4-4} = 2$.The inverse point $Q(a, b)$ of $P(3, 3)$ lies on the line $OP$. The slope of $OP$ is $m = \frac{3 - (-2)}{3 - 2} = \frac{5}{1} = 5$.The equation of line $OP$ is $y - (-2) = 5(x - 2)$,which simplifies to $y = 5x - 12$.Since $Q(a, b)$ lies on this line,$b = 5a - 12$ (Equation $1$).By the property of inverse points,$O, Q, P$ are collinear and $OQ \cdot OP = r^2$.$OP = \sqrt{(3-2)^2 + (3-(-2))^2} = \sqrt{1^2 + 5^2} = \sqrt{26}$.$OQ = \sqrt{(a-2)^2 + (b+2)^2} = \sqrt{(a-2)^2 + (5a-12+2)^2} = \sqrt{(a-2)^2 + (5a-10)^2} = \sqrt{(a-2)^2 + 25(a-2)^2} = \sqrt{26(a-2)^2} = |a-2|\sqrt{26}$.Given $OQ \cdot OP = r^2$,we have $|a-2|\sqrt{26} \cdot \sqrt{26} = 2^2 = 4$.$26|a-2| = 4 \Rightarrow |a-2| = \frac{4}{26} = \frac{2}{13}$.Since $Q$ lies on the same side of $O$ as $P$,$a-2 = \frac{2}{13} \Rightarrow a = 2 + \frac{2}{13} = \frac{28}{13}$.Then $b = 5(\frac{28}{13}) - 12 = \frac{140 - 156}{13} = -\frac{16}{13}$.Finally,$a+5b = \frac{28}{13} + 5(-\frac{16}{13}) = \frac{28 - 80}{13} = -\frac{52}{13} = -4$.

If the inverse point of the point $P(3, 3)$ with respect to the circle $x^2+y^2-4x+4y+4=0$ is $Q(a, b)$,then $a+5b=$

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