The equation of a circle is given by \(x^{2} + y^{2} - 4x - 2y - 3\). Find the radius and the coordinates of its centre.
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Correct Answer: Option C
Explanation:
Equation of a circle with radius r and centre (a, b).
= \((x - a)^{2} + (y - b)^{2} = r^{2}\)
Expanding, we have
\(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)
Comparing, with \(x^{2} + y^{2} - 4x - 2y - 3 = 0\)
\(2a = 4 \implies a = 2\)
\(2b = 2 \implies b = 1\)
\(r^{2} - a^{2} - b^{2} = 3 \implies r^{2} = 3 + 2^{2} + 1^{2} = 8\)
\(r = 2\sqrt{2}\)
Equation of a circle with radius r and centre (a, b).
= \((x - a)^{2} + (y - b)^{2} = r^{2}\)
Expanding, we have
\(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)
Comparing, with \(x^{2} + y^{2} - 4x - 2y - 3 = 0\)
\(2a = 4 \implies a = 2\)
\(2b = 2 \implies b = 1\)
\(r^{2} - a^{2} - b^{2} = 3 \implies r^{2} = 3 + 2^{2} + 1^{2} = 8\)
\(r = 2\sqrt{2}\)