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Image of the circle \mathrm{x^{2}+y^{2}-6 x+8=0} in the line \mathrm{y=x} is

Option: 1

\mathrm{x^{2}+y^{2}+6 y+8=0}


Option: 2

\mathrm{x^{2}+y^{2}-6 y-8=0}


Option: 3

\mathrm{x^{2}+y^{2}-6 y+8=0}


Option: 4

None of these.


Answers (1)

best_answer

Centre and radius of the circle \mathrm{x^{2}+y^{2}-6 x+8=0} is (3,0) and 1.

Image of centre (3,0)  in the line  \mathrm{x-y=0} is (0,3).

Therefore, image of the given circle in the line y=x is

\mathrm{(x-2)^{2}+(y-3)^{2}=1 \text { P } x^{2}+y^{2}-6 y+8=0}.

Hence (C) is the correct answer.

 

Posted by

Sanket Gandhi

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