Question 30361
 x^2+y^2-6x-2y-6=0 

A circle, of radius r, and centred on the point (h,k) can be written as,
(x-h)² + (y-k)² = r²
expanding this expression,
x² - 2hx + h² + y² - 2ky + k² = r²
x² + y² - 2hx - 2ky + (h² + k² - r²) = 0
comparing this with the expression we were given,
x^2+y^2-6x-2y-6=0 
and comparing coefficients,
-2h = -6
-2k = -2
h² + k² - r² = -6
From the 1st two eqns,
h = 3
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k = 1
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substituting for h=3 and k=1 into the 3rd eqn,
9 + 1 - r² = -6
-r² = -16
r² = 16
r = 4
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We therefore have: (h,k) = (3,1) and r = 4
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Ans:C
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