Parametric equation of a circle

In this article, we will cover the concept of the parametric form of a circle. This concept falls under the broader category of coordinate geometry. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of two questions have been asked on this concept, including two in 2021.

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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 .