The centre of a circle is \((2,-3)\) and the circumference is \(10 \pi\). Then its equation is
The centre of a circle is \((2,-3)\) and the circumference is \(10 \pi\). Then its equation is
\(x^2+y^2+4 x+6 y+12=0\)
\(x^2+y^2-4 x+6 y+12=0\)
\(x^2+y^2-4 x+6 y-12=0\)
\(x^2+y^2-4 x-6 y-12=0\)
Solution
Let the radius of required circle is ' \(r\) ', so the circumference \(=2 \pi r=10 \pi\) (given)
\(\Rightarrow \quad r=5\)
and centre of required circle is \((2,-3)\)
So, the equation of required circle is
\(\begin{aligned}
(x-2)^2+(y+3)^2 & =25 \\
\Rightarrow \quad x^2+y^2-4 x+6 y-12 & =0
\end{aligned}\)
Hence, option (c) is correct.