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Question

In the equation (xāˆ’3)2+(yāˆ’2)2=16, the center of the circle is

A
(3,2)
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B
(3,2)
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C
(3,2)
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D
(2,3)
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Solution

The correct option is A (3,2)
(x3)2+(y2)2=16(i)
General standard equation of circlex2+y2=r2(ii)
Comparing (i) and (ii), (ii) have centre at (0,0). In (i) only the origin is shifted. So, centre of (i) (h,k):(3,2)

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