Question

# The parametric form of equation of the circle ${{x}^{2}}+{{y}^{2}}-6x+2y-28=0$ isA. $x=-3+\sqrt{38}\cos \theta ,y=-1+\sqrt{38}\sin \theta$ B. $x=\sqrt{28}\cos \theta ,y=\sqrt{28}\sin \theta$C. $x=-3-\sqrt{38}\cos \theta ,y=1+\sqrt{38}\sin \theta$D. $x=3+\sqrt{38}\cos \theta ,y=-1+\sqrt{38}\sin \theta$E. $x=3+38\cos \theta ,y=-1+38\sin \theta$

Hint: We have to define a circle with the locus of all points that satisfy the equations.
$x=r\cos \theta \ and\ y=r\sin \theta$
Where x, y are the coordinates of any point on the circle, â€˜râ€™ is the radius of the circle. Also, $\theta$ is the parameter of the angle subtended by the point at the circleâ€™s centre.

Given the equation of a circle is,
${{x}^{2}}+{{y}^{2}}-6x+2y-28=0$
We have to find the parametric equation of the circle.
We know that the parametric equation of a circle can be defined as the locus of all points that satisfy the equations.
$x=r\cos \theta \ and\ y=r\sin \theta$
Where â€˜râ€™ is the radius of the circle and (x, y) are the coordinates of any point on the circle.
$\theta$ is the parameter of the angle subtended by the point at the centre of the circle.
A figure representing the above details can be drawn as shown below:

To find the parametric form of a circle, first we have to find the coordinate of centre and the radius of the circle from the given equation of circle.
We have,
\begin{align} Â Â & {{x}^{2}}+{{y}^{2}}-6x+2y-28=0 \\ Â & \Rightarrow \left( {{x}^{2}}-6x \right)+\left( {{y}^{2}}+2y \right)-28=0 \\ \end{align}
Adding and subtracting 3 and 1 to complete the squares, we get,
\begin{align} Â Â & \Rightarrow {{\left( x \right)}^{2}}+{{\left( 3 \right)}^{2}}-2\left( 3 \right)\left( x \right)-9+{{\left( y \right)}^{2}}+2\left( y \right)\left( 1 \right)+{{\left( 1 \right)}^{2}}-1-28=0 \\ Â & \Rightarrow {{\left( x-3 \right)}^{2}}-9+{{\left( y+1 \right)}^{2}}-1-28=0 \\ Â & \Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{\left( \sqrt{38} \right)}^{2}}..................\left( 1 \right) \\ \end{align}
We know that the general equation of the circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{\left( r \right)}^{2}}$.
Where, (h, k) are coordinates of the centre of the circle and r is the radius of the circle.
We know that the parametric equation of the circle is,
\begin{align} Â Â & x=a+r\cos \theta \\ Â & y=b+r\sin \theta \\ \end{align}
Here, (a, b) are coordinates of the centre of the circle.
And we have coordinates of the centre of the circle is (3, -1) and radius $=\sqrt{38}$ .
Putting the value of coordinates of centre and radius in general parametric equation, we get,
\begin{align} Â Â & x=3+\sqrt{38}\cos \theta \\ Â & y=-1+\sqrt{38}\sin \theta \\ \end{align}
Hence option D is the correct answer.

Note: Always find the coordinates (h, k) of the centre of the circle and radius â€˜râ€™ by converting the given equation to the general equation of the circle.
General equation of the circle:
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{\left( r \right)}^{2}}$