Parametric equation of a circle

Parametric Form

To represent any point on a curve in terms of a single variable (parameter), we use parametric form of that curve.

1. Parametric Form for $x^2+y^2=r^2$
$P(x, y)$ is a point on the circle $x^2+y^2=r^2$ with centre $O(0,0)$. And $O P$ makes an angle $\theta$ with the positive direction of the $X$-axis, then $x=r \cdot \cos \theta$, $y=r \cdot \sin \theta$ called the parametric equation of the circle.
Here as $\theta$ varies, the point on the circle also changes, and thus $\theta$ is called the parameter. Here $0 \leq \theta<2 \pi$.
So any arbitrary point on this circle can be assumed as $(r \cdot \cos \theta, r \cdot \sin \theta)$

2. Parametric Form for $(x-h)^2+(y-k)^2=r^2$

Centre of the circle here is $(\mathrm{h}, \mathrm{k})$.
Parametric point on it is $(h+r \cdot \cos \theta, k+r \cdot \sin \theta)$.

Recommended Video Based on Parametric Form of Circles

Solved Examples Based On Parametric Form of Circles:

Example 1: Let $\mathbf{A}(1,4)$ and $\mathbf{B}(1,-5)$ be two points. Let $\mathbf{P}$ be a point on the circle $(x-1)^2+(y-1)^2=1$ such that has maximum value, then the points, $\mathbf{P}, \mathbf{A}$, and $\mathbf{B}$ lie on:
1) a parabola
2) a straight line
3) a hyperbola
4) an ellipse

Solution:
Let P be any point on the circle $(x-1)^2+(y-1)^2=1$
So, $\mathrm{P}=(1+\cos \theta, 1+\sin \theta)$
$A(1,4), B(1,-5)$
$(P A)^2+(P B)^2$
$=\left((1+\cos \theta-1)^2+(1+\sin \theta-4)^2\right)+\left((1+\cos \theta-1)^2+(1+\sin \theta+5)^2\right)$
$=\cos ^2 \theta+\sin ^2 \theta-6 \sin \theta+9+\cos ^2 \theta+\sin ^2 \theta+12 \sin \theta+36$
$
=47+6 \sin \theta
$
is maximum if $\sin \theta=1$
$
\begin{aligned}
& \Rightarrow \sin \theta=1, \cos \theta=0 \\
& \mathrm{P}(1,1), \mathrm{A}(1,4), \mathrm{B}(1,-5)
\end{aligned}
$
$P, A$, and $B$ lie in a straight line.
Hence, the answer is the option 2.

Example 2: What are the parametric coordinates of $x^2+y^2=36$ ?
Solution:
As we have learned
Parametric form for $x^2+y^2=r^2$ is
$
x=r \cos \theta, y=r \sin \theta
$

Since, in $x^2+y^2=36$
$
r=6
$

Thus $x=6 \cos \theta, y=6 \sin \theta$

Example 3: Which of the following is true for the Circles $C_1: x^2+y^2-2 x+4 y=8$ and $C_2: a x^2+a y^2-2 a x+4 a y=8$ where $a>0$ ?
1) $C_1$ and $C_2$ are concentric circles for all values of a.
2) $C_1$ is lying entirely inside $C_2$ if $0<a<1$.
3) $C_2$ is lying entirely inside $C_1$ if $a>1$.
4) All of the above

Solution:
$C_1: x^2+y^2-2 x+4 y=8$
$C_2: x^2+y^2-2 x+4 y=\frac{8}{a}$
$C_1$ and $C_2$ both have same centre $(1,-2)$
So they are concentric for any value of a.
Now radius for the first circle $=\sqrt{1+4+8}=\sqrt{13}$
The radius for the second circle $=\sqrt{1+4+\frac{8}{a}}=\sqrt{5+\frac{8}{a}}$
- For $0<\mathrm{a}<1$
radius of the second circle > radius of the first circle
As they are concentric, so in this case the first circle lies entirely inside the second circle
- For $a>1$

radius of second circle < radius of first circle

As they are concentric, so in this case, the second circle lies entirely inside the first circle

Hence, the answer is the option 4.

Example 4: What are the parametric coordinates for $(x-5)^5+(y+2)^2=49$ ?
Solution:
As we have learned
Parametric form:
$
\begin{aligned}
& x=h+r \cos \theta \\
& y=k+r \sin \theta
\end{aligned}
$

For a circle having centre $(h, k)$ and radius $r$.
$
\begin{gathered}
x-5=7 \cos \theta \Rightarrow x=7 \cos \theta+5 \\
y+2=7 \sin \theta \Rightarrow y=7 \sin \theta-2
\end{gathered}
$

Grample 5: Two circles are inscribed and circumscribed about a square A B C D, the length of each side of the square is 16 . $P$ and $Q$ are two points respectively on these circles, then $\Sigma(P A)^2+\Sigma(Q A)^2$ is equal to

Solution:
Let the center of the square be the origin $O$ and the lines through $O$ parallel to the sides of the square be the coordinate axis. Then the vertices of the square are $A(8,8), B(-8,8), C(-8,-8)$ and $D(8,-8)$
The radii of the inscribed and circumscribed circles are respectively 8 and $O A=\sqrt{\left(8^2+8^2\right)}=8 \sqrt{2}$ and their center is at the origin.
Let the coordinate of $P$ be the $(8 \cos \theta, 8 \sin \theta)$ and that of $Q b e(8 \sqrt{2} \cos \phi, 8 \sqrt{2} \sin \phi)$
Then $\Sigma(P A)^2=8^2\left[(\cos \theta-1)^2+(\sin \theta-1)^2+(\cos \theta+1)^2+(\sin \theta-1)^2\right.$
$\left.+(\cos \theta+1)^2+(\sin \theta+1)^2-(\cos \theta-1)^2+(\sin \theta+1)^2\right]=2 \times 8^2\left[2 \cos ^2 \theta+2+2 \sin ^2 \theta+2\right]$
$
=12 \times 8^2
$

Similarly $\Sigma(Q A)^2=16 \times 8^2$
$
\therefore \Sigma(P A)^2+\Sigma(Q A)^2=28 \times 8^2=1792
$

Hence, the answer is 1792 .