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Question

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


A
(0, 0)
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B
(1, 0)
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C
(0, 1)
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D
(1, 1)
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Solution

The correct option is C (0, 1)
$$eq^{n}$$ 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 $$x^{2}+y^{2}-4x-6y+9=0$$
$$PO\times OA =r^{2} =(2)^{2}=4$$
$$PO=\frac{4}{\sqrt{2}}=2\sqrt{2}$$
$$\therefore (h-2)^{2}+(h-2)^{2}=(2\sqrt{2})^{2}$$
$$2(h-2)^{2}=8$$
(h-2)=2   or h-2=-2
h=4    h=0
than k=5   or  k=1
$$\Rightarrow p(4,5) \quad  or \quad p(0,1)  $$
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)


57882_33445_ans_6e2fbffe20ed4c008c8115b923dc955d.png

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