Find the equation of a circle with centre (2, -3) and radius 2 units.
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Correct Answer: Option A
Explanation:
The equation of a circle with centre coordinate (a, b) and radius r is :
\((x - a)^{2} + (y - b)^{2} = r^{2}\)
Given centre = (2, -3) and radius r = 2 units
Equation = \((x - 2)^{2} + (y - (-3))^{2} = 2^{2}\)
\(x^{2} - 4x + 4 + y^{2} + 6y + 9 = 4\)
\(x^{2} + y^{2} - 4x + 6y + 4 + 9 - 4 = 0 \implies x^{2} + y^{2} - 4x + 6y + 9 = 0\)
The equation of a circle with centre coordinate (a, b) and radius r is :
\((x - a)^{2} + (y - b)^{2} = r^{2}\)
Given centre = (2, -3) and radius r = 2 units
Equation = \((x - 2)^{2} + (y - (-3))^{2} = 2^{2}\)
\(x^{2} - 4x + 4 + y^{2} + 6y + 9 = 4\)
\(x^{2} + y^{2} - 4x + 6y + 4 + 9 - 4 = 0 \implies x^{2} + y^{2} - 4x + 6y + 9 = 0\)