Answer

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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, ‘r’ is the radius of the circle. Also, $\theta $ is the parameter of the angle subtended by the point at the circle’s centre.

Complete step-by-step answer:

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 ‘r’ 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}

& {{x}^{2}}+{{y}^{2}}-6x+2y-28=0 \\

& \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}

& \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 \\

& \Rightarrow {{\left( x-3 \right)}^{2}}-9+{{\left( y+1 \right)}^{2}}-1-28=0 \\

& \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}

& x=a+r\cos \theta \\

& 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}

& x=3+\sqrt{38}\cos \theta \\

& 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 ‘r’ by converting the given equation to the general equation of the circle.

General equation of the circle:

${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{\left( r \right)}^{2}}$

$x=r\cos \theta \ and\ y=r\sin \theta $

Where x, y are the coordinates of any point on the circle, ‘r’ is the radius of the circle. Also, $\theta $ is the parameter of the angle subtended by the point at the circle’s centre.

Complete step-by-step answer:

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 ‘r’ 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}

& {{x}^{2}}+{{y}^{2}}-6x+2y-28=0 \\

& \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}

& \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 \\

& \Rightarrow {{\left( x-3 \right)}^{2}}-9+{{\left( y+1 \right)}^{2}}-1-28=0 \\

& \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}

& x=a+r\cos \theta \\

& 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}

& x=3+\sqrt{38}\cos \theta \\

& 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 ‘r’ by converting the given equation to the general equation of the circle.

General equation of the circle:

${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{\left( r \right)}^{2}}$

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