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Equation: x^2 +y^2 -4x -6y -3 = 0

Completing the squares: (x-2)^2 +(y-3)^2 -4 -9 -3 = 0

So: (x-2)^2 +(y-3) = 16

Therefore the centre of the circle is at (2, 3) and its radius is 4

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Q: What is the centre and radius of a circle whose equation is x squared plus y squared -4x -6y -3 equals 0 showing work?
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