The image of the point (3,5) in the line x–y+1=0, lies on :
A
(x−4)2+(y−4)2=8
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B
(x−2)2+(y−4)2=4
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C
(x−2)2+(y−2)2=12
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D
(x−4)2+(y+2)2=16
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Solution
The correct option is B(x−2)2+(y−4)2=4 Image of P(3,5) on the line x–y+1=0 is x−31=y−5−1=−2(3−5+1)2=1 x=4,y=4 ∴ Image is (4,4)
Which lies on (x−2)2+(y−4)2=4