The radius of the circle having centre at (2,1) and whoseone of the chord is diameter of the circle x 2 + y 2 – 2x – 6y + 6 =0 is??Sir please explain!!

Question

Sunil Raikwar · Answer

let the radius of circle be r then equation of circle
(x-2)²+(y-1)²=r²
x²+y²-4x-2y+5-r²=0...................1
x²+y²-2x-6y+6=0........................2
common chord of circle 1 & 2
s₁-s₂=0
2x-4y+1+r²=0
this passes through (1,3)
2-12+1+r²=0
r²=9
r=3

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The radius of the circle having centre at (2,1) and whoseone of the chord is diameter of the circle x 2 + y 2 – 2x – 6y + 6 =0 is??Sir please explain!!

The radius of the circle having centre at (2,1) and whoseone of the chord is diameter of the circle x² + y² – 2x – 6y + 6 =0 is??Sir please explain!!

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The radius of the circle having centre at (2,1) and whoseone of the chord is diameter of the circle x 2 + y 2 – 2x – 6y + 6 =0 is??Sir please explain!!

The radius of the circle having centre at (2,1) and whoseone of the chord is diameter of the circle x2 + y2 – 2x – 6y + 6 =0 is??Sir please explain!!

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The radius of the circle having centre at (2,1) and whoseone of the chord is diameter of the circle x² + y² – 2x – 6y + 6 =0 is??Sir please explain!!