Question

The center of a circle is $\left( {2, - 3} \right)$and the 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

Hint: We are given the circumference of the circle. We can find the radius from the circumference using the equation $C = 2\pi r$ Then we can find the equation of the circle by using the center and radius in the equation ${\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {r^2}$. After simplification, we will get the required equation.

Complete step by step answer:

We have circumference as $10\pi $. We know that circumference is given by $2\pi r$.
\[\therefore 2\pi r = 10\pi \]
$ \Rightarrow r = \dfrac{{10\pi }}{{2\pi }} = 5$
We have center of the circle as $\left( {2, - 3} \right)$ and radius is 5.
Equation of circle with radius r and centre $({x_0},{y_0})$ is given by ${\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {r^2}$.
Then the equation of circle is
\[ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y + 3} \right)^2} = {5^2}\]
On expanding the squares using the algebraic identity ${\left( {a \pm b} \right)^2} = {a^2} \pm 2ab + {b^2}$ we get,
\[ \Rightarrow {x^2} - 4x + 4 + {y^2} + 6y + 9 = 25\]
By rearranging, we get.
\[
   \Rightarrow {x^2} + {y^2} - 4x + 6y + 13 - 25 = 0 \\
   \Rightarrow {x^2} + {y^2} - 4x + 6y - 12 = 0 \\
 \]
So, the required equation of the circle is \[{x^2} + {y^2} - 4x + 6y - 12 = 0\]
Therefore, the correct answer is option C.

Note: The main concept used here is the equation of circles. Equation of circle with radius r and centre $({x_0},{y_0})$ is given by ${\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {r^2}$. The simple algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ is used to simplify the equation of the circle to required form. A circle can be defined as the collection of all the points that are equidistant from its center. This distance is called the radius of the circle. Every point $(x,y)$ that satisfy the equation of the circle lie on the circle. If the point $(x,y)$ is the inside the circle, then the value of the equation of the circle at $(x,y)$ will be less than 0 and value of the equation of the circle at $(x,y)$ will be greater than 0 if the point $(x,y)$ is the outside the circle. We must take care of the signs of the coordinates of the centre will substituting in the equation of the circle. We must also take care of the sign while squaring using the identity.