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Question

 The image of the circle $$\mathrm{x}^{2}+\mathrm{y}^{2}-2\mathrm{x}=0$$ in the line $$\mathrm{x}+\mathrm{y}=2$$ is the circle


A
x2+y24x2y+4=0
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B
x2+y2+2x=0
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C
x2+y22y=0
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D
x2+y22x2y+4=0
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Solution

The correct option is A $$\mathrm{x}^{2}+\mathrm{y}^{2}-4\mathrm{x}-2\mathrm{y}+4=0$$
$$x^2+y^2-2x=0$$
centre $$ (1,0)$$   radius $$=1 = r$$
So, image of centre $$(1,0)$$ w.r.t. $$(x+y=2)$$ is
$$\Rightarrow \dfrac{h-1}{1}=\dfrac{k-0}{1}=-2\dfrac{(-1)}{1+1}=1$$
$$h=2$$  and $$k=1$$
$$(2,1)$$ is image of $$(1,0)$$
So, $$eq^n$$ of image circle is
$$(x-2)^+(y-1)^2=r^2=1$$
$$\Rightarrow x^2+y^2-4x-2y+4=0$$

Mathematics

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