Find the equation of the circle with centre $(1, -2)$ and radius $3$.
A$x^2 + y^2 - 2x + 4y - 4 = 0$CORRECT
B$x^2 + y^2 - 2x + 4y + 2 = 0$
C$x^2 + y^2 + 2x - 4y - 4 = 0$
D$x^2 + y^2 - x + 2y - 9 = 0$
JUPEBMathematicsCoordinate Geometry
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Why the answer is A, and why the others tempt you.
$(x - 1)^2 + (y + 2)^2 = 9$ expands to $x^2 - 2x + 1 + y^2 + 4y + 4 = 9$, which simplifies to $x^2 + y^2 - 2x + 4y - 4 = 0$. Using $+2$ instead of $-4$ comes from forgetting to move the 9 across or from squaring the radius incorrectly.
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