Question

How do you write the equation of the circle with center $\left( 0,0 \right)$, radius $=6$?

Answer 1

Hint: In order to write the equation of a circle, the two basic things that we must know are its center and its radius, hence we are provided with this data in the problem above. Now, we shall use the standard form of the equation of the circle to find its equation. This form is different from the general form of the equation of circle and makes the calculations much simpler.

Complete answer:
In order to find the solution of this question, we should have knowledge of the standard equation of a circle when radius and center are given to us. Because we have been given that the radius of the circle is 6 and the center of the circle is $\left( 0,0 \right)$.
The standard form of equation of a circle with center $\left( h,k \right)$ and radius $r$ is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.
When the center of the circle is the origin, that is, the x-coordinate of center of circle, $h=0$ and the y-coordinate of center of circle, $k=0$, then the equation transforms as:
\[{{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{r}^{2}}\]
$\Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}$
Now, we shall substitute the value of the radius of the circle, $r=6$, to find the complete equation:
 $\Rightarrow {{x}^{2}}+{{y}^{2}}={{\left( 6 \right)}^{2}}$
And we know that ${{6}^{2}}=36$. Therefore, we get,
$\Rightarrow {{x}^{2}}+{{y}^{2}}=36$
Therefore, the equation of the given circle is ${{x}^{2}}+{{y}^{2}}=36$.

Note: The general equation of any curve is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+2hxy+c=0$
For a circle, $h=0$, $x$ is replaced with $\dfrac{\left( x+{{x}_{1}} \right)}{2}$ and $y$ is replaced with $\dfrac{\left( y+{{y}_{1}} \right)}{2}$,
Where, $\left( {{x}_{1}}+{{y}_{1}} \right)$ is the center of the circle
Thus, the equation is transformed as:
\[{{x}^{2}}+{{y}^{2}}+2g\left( \dfrac{x+{{x}_{1}}}{2} \right)+2f\left( \dfrac{y+{{y}_{1}}}{2} \right)+c=0\]
The equation of circle also has a parametric form. The parameter in such an equation is $\theta $, which is the angle made by the radius vector from a particular point on the circle with the positive x-axis. The parametric point is$\left( r\cos \theta ,r\sin \theta \right)$ which is dependent on the radius, $r$ and the angle, $\theta $. When we substitute these points in the standard equation of the circle, we get the parametric equation of the circle.