If Q is the inverse point of the point P(2, 3 | vedclass

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(C) The given circle is $x^2+y^2-2x-2y+1=0$,which can be written as $(x-1)^2+(y-1)^2=1$. The center is $O(1, 1)$ and radius $r=1$.Let $Q$ be the inver

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$5x^2+5y^2-16x-22y+33=0$

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(C) The given circle is $x^2+y^2-2x-2y+1=0$,which can be written as $(x-1)^2+(y-1)^2=1$. The center is $O(1, 1)$ and radius $r=1$.Let $Q$ be the inverse point of $P(2, 3)$. The inverse point $Q$ lies on the line $OP$ such that $OP 	imes OQ = r^2$.The slope of $OP$ is $\frac{3-1}{2-1} = 2$. The equation of line $OP$ is $y-1 = 2(x-1)$,which simplifies to $2x-y-1=0$.$OP = \sqrt{(2-1)^2+(3-1)^2} = \sqrt{1+4} = \sqrt{5}$.Since $OP 	imes OQ = r^2 = 1$,we have $OQ = \frac{1}{\sqrt{5}}$.Let $Q$ be $(h, k)$. Since $Q$ lies on $2x-y-1=0$,$k = 2h-1$. Also,the distance $OQ = \sqrt{(h-1)^2+(k-1)^2} = \frac{1}{\sqrt{5}}$.Substituting $k-1 = 2h-2$,we get $\sqrt{(h-1)^2+(2h-2)^2} = \frac{1}{\sqrt{5}}$ $\Rightarrow \sqrt{5(h-1)^2} = \frac{1}{\sqrt{5}}$ $\Rightarrow 5(h-1)^2 = 1$ $\Rightarrow (h-1)^2 = \frac{1}{5}$.$h-1 = \pm \frac{1}{\sqrt{5}} \Rightarrow h = 1 \pm \frac{1}{\sqrt{5}}$.For $h = 1 + \frac{1}{\sqrt{5}}$,$k = 2(1 + \frac{1}{\sqrt{5}}) - 1 = 1 + \frac{2}{\sqrt{5}}$.Thus $Q = (1 + \frac{1}{\sqrt{5}}, 1 + \frac{2}{\sqrt{5}})$.The equation of the circle with diameter $PQ$ is $(x-2)(x-(1+\frac{1}{\sqrt{5}})) + (y-3)(y-(1+\frac{2}{\sqrt{5}})) = 0$.Expanding this,we get $x^2 - (3+\frac{1}{\sqrt{5}})x + 2(1+\frac{1}{\sqrt{5}}) + y^2 - (4+\frac{2}{\sqrt{5}})y + 3(1+\frac{2}{\sqrt{5}}) = 0$.$x^2+y^2 - (3+\frac{1}{\sqrt{5}})x - (4+\frac{2}{\sqrt{5}})y + 5 + \frac{8}{\sqrt{5}} = 0$.Multiplying by $5$ and simplifying,we obtain $5x^2+5y^2-16x-22y+33=0$.

If $Q$ is the inverse point of the point $P(2, 3)$ with respect to the circle $x^2+y^2-2x-2y+1=0$,then the circle with $PQ$ as diameter is

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