Question

The centroid of a circle is $\left( {2, - 3} \right)$ and circumference is $10\pi$ .Then the equation of the circle is
A.${x^2} + {y^2} + 4x + 6y + 12 = 0$
B.${x^2} + {y^2} - 4x + 6y + 12 = 0$
C.${x^2} + {y^2} - 4x + 6y - 12 = 0$
D.${x^2} + {y^2} - 4x - 6y - 12 = 0$

Answer 1

We are given with the center of the circle and its circumference. From this we will find the radius of the circle using the formula $2\pi r$ . Then using the general equation of circle ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ and putting the value of center of circle we will get equation of circle.

Complete step-by-step answer:
Given that, circumference of a circle is $10\pi$
$\Rightarrow 10\pi = 2\pi r$
Cancelling $\pi$ from both sides,
$\Rightarrow r = 5unit.$
Now we know that the general form of the equation is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$.
Center of the circle is $\left( {h,k} \right) = \left( {2, - 3} \right)$ and radius $r = 5$.
Putting these values in the equation above
$\Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - \left( { - 3} \right)} \right)^2} = {5^2}$
Performing the expansions using the identity
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$

$$\Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - \left( { - 3} \right)} \right)^2} = {5^2} \\\ \Rightarrow {x^2} - 4x + 4 + \left( {{y^2} + 6y + 9} \right) = 25 \\\ \Rightarrow {x^2} + {y^2} - 4x + 6y + 4 + 9 = 25 \\\ \Rightarrow {x^2} + {y^2} - 4x + 6y + 13 = 25 \\\ \Rightarrow {x^2} + {y^2} - 4x + 6y = 25 - 13 \\\ \Rightarrow {x^2} + {y^2} - 4x + 6y = 12 \\\$$

And this is the equation of the circle ${x^2} + {y^2} - 4x + 6y = 12$.
Hence option B is correct.

Note: We are given with four options here with slight difference in the signs only. So be careful when you expand the brackets and add or subtract the terms. Because a minor negligence will make your answer wrong.