Find the centre and radius of the circle $x^2 + y^2 - 4x + 6y - 12 = 0$.
Acentre $(-2, 3)$, radius $5$
Bcentre $(4, -6)$, radius $5$
Ccentre $(2, -3)$, radius $\sqrt{12}$
Dcentre $(2, -3)$, radius $5$CORRECT
JUPEBMathematicsCoordinate Geometry
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Why the answer is D, and why the others tempt you.
Complete the square: $(x - 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0$, so $(x - 2)^2 + (y + 3)^2 = 25$. The centre is $(2, -3)$ and the radius is $\sqrt{25} = 5$. Taking the centre as $(-2, 3)$ reverses the signs, and $\sqrt{12}$ ignores the completed-square constants.
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