The inverse point of (1, 2) with respect to the circle $x^{2} + y^{2} - 4 x - 6 y + 9 = 0$ is

(0, 0)

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(1, 0)

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(0, 1)

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(1, 1)

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Solution

The correct option is C (0, 1)
$e q^{n}$ OA is
y-2=1(x-1)
y-x=1-----(1)
then p (h,k)= p (h,h+1)
given p is inverse point of (1,2) w.rt cirde $x^{2} + y^{2} - 4 x - 6 y + 9 = 0$
$P O \times O A = r^{2} = (2)^{2} = 4$
$P O = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$
$∴ (h - 2)^{2} + (h - 2)^{2} = (2 \sqrt{2})^{2}$
$2 (h - 2)^{2} = 8$
(h-2)=2 or h-2=-2
h=4 h=0
than k=5 or k=1
$\Rightarrow p (4, 5) o r p (0, 1)$
But p and Q (1,2) lies on some side of centre of arde
$\Rightarrow p = (0, 1)$ is inverse of point of (1,2)

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The inverse point of (1, 2) with respect to the circle $x^{2} + y^{2} - 4 x - 6 y + 9 = 0$ is