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Question

A circle of radius 8 is passing through origin and the point (4,0). If the centre lies on the line y=x, then the equation of the circle is

A
(x2)2+(y2)2=8
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B
(x+2)2+(y+2)2=8
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C
(x3)2+(y3)2=8
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D
(x+3)2+(y+3)2=8
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E
(x4)2+(y4)2=8
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Solution

The correct option is A (x2)2+(y2)2=8
Let the equation of the circle be (xa)2+(ya)2=8[Centre lies on x=yand radius is 8]

Both (0,0) and (4,0) lies of the circle, therefore we have the below 2 equations

a2+a2=8a=±2

(4a)2+a2=8a24a+4=0(a2)2=0a=2

,a=2 and the required equation is (x2)2+(y2)2=8

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