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Question

# If one of the diameters of the circle x2+y2−2x−6y+6=0 is a chord to the circle with centre (2,1), then the equation of the circle is

A
x2+y24x2y+2=0
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B
x2+y24x2y2=0
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C
x2+y24x2y4=0
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D
x2+y24x2y+4=0
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Solution

## The correct option is C x2+y2−4x−2y−4=0According to the given question The given circle is x2+y2−2x−6y+6=0 C=(−g,−f)=(1,3) r=√g2+f2−c=√1+9−6=2 Let AB be one of its diameter which is the chord of other circle with center C1(2,1), so in △C1CB, we have (C1B)2=(CC1)2+(CB)2⇒r2=[(2−1)2+(1−3)2]+22 ⇒r=3 Hence, the equation of the other circle is (x−2)2+(y−1)2=32⇒x2+y2−4x−2y−4=0

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