Circles in Maths: Definition, Formulas, Properties and Examples
In this article, we will cover the concept of the 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 twenty-one questions have been asked on this concept, including three in 2013, one in 2014, one in 2015, two in 2016, one in 2017, one in 2018, six in 2019, two in 2020, two in 2021, and two in 2022.
Definition of Cirlce
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A circle is the locus of a moving point such that its distance from a fixed point is constant.
The fixed point is called the centre ( O ) of the circle and the constant distance is called its radius $(r)$.
Equation of circle
Centre-Radius Form
The equation of a circle with centre at $C(h, k)$ and dadius $r$ is $(x-h)^2+(y-k)^2=r^2$
Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be any point on the circle. Then, by definition, $|C P|=r$.
Using the distance formula, we have
$
\sqrt{(x-h)^2+(y-k)^2}=r
$
i.e.
$
(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2
$
If the centre of the circle is the origin or $(0,0)$ then the equation of the circle becomes
$
\begin{aligned}
& (x-0)^2+(y-0)^2=r^2 \\
& \text { i.e. } x^2+y^2=r^2
\end{aligned}
$
General Form
The equation of a circle with centre at $(\mathrm{h}, \mathrm{k})$ and radius r is
$
\begin{aligned}
& \Rightarrow(x-h)^2+(y-k)^2=r^2 \\
& \Rightarrow x^2+y^2-2 h x-2 k y+h^2+k^2-r^2=0
\end{aligned}
$
Which is of the form :
$
x^2+y^2+2 g x+2 f y+c=0
$
This is known as the general equation of the circle.
To get the radius and centre if only the equation of the circle (ii) is given:
Compare eq (i) and eq (ii)
$
\mathrm{h}=-\mathrm{g}, \mathrm{k}=-\mathrm{h} \text { and } \mathrm{c}=\mathrm{h}^2+\mathrm{k}^2-\mathrm{r}^2
$
Coordinates of the centre $(-\mathrm{g},-\mathrm{f})$
$
\text { Radius }=\sqrt{g^2+f^2-c}
$
Nature of the Circle
For the standard equation of a circle $\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{~g} x+2 \mathrm{fy}+\mathrm{c}=0$ whose radius is given as $\sqrt{g^2+f^2-c}$
Now the following cases arise
1. If $\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}>0$, then the radius of the circle will be real. Hence, the circle is a real circle.
2. If $\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}=0$, then the radius of the circle will be real $(=0)$. Hence, the circle is a Point circle because the radius is 0 .
3. If $\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}<0$, then the radius of the circle will be imaginary. Hence, the circle is an imaginary circle.
Recommended Video Based on Circles
Solved Examples Based on Circles
Example 1: A circle with an equation $x^2+y^2+4 x+2 f y+c=0$ has centre $(-2,6)$ and radius $=\sqrt{7}$. Find c .
Solution:
As we learned
Centre of a circle and radius -
The fixed point in the circle is called the centre and the fixed distance is called the radius.
Here, $g=-2$ and $f=-6$,
$
\begin{gathered}
\qquad \sqrt{g^2+f^2-c}=\sqrt{7} \\
\Rightarrow 4+36-c-7 \\
\text { Thus } \Rightarrow c=33
\end{gathered}
$
Hence, the answer is 33.
Example 2: The axes are translated so that the new equation of the circle $x^2+y^2-5 x+2 y-5=0$ has no first degree terms. Then the new equation is:
Solution:
Equation of a circle -
$
x^2+y^2=r^2
$
- wherein
Circle with centre $(O, O)$ and radius $r$.
Equation of a circle -
$
(x-h)^2+(y-k)^2=r^2
$
- wherein
Circle with centre $(h, k)$ and radius $r$.
$
\begin{aligned}
& x^2+y^2-5 x+2 y-5=0 \\
& (x-5 / 2)^2+(y+1)^2-5-25 / 4-1=0 \\
& (x-5 / 2)^2+(y+1)^2=49 / 4
\end{aligned}
$
so the axes are shifted to $(5 / 2,-1)$ New equation of circle must be
$
x^2+y^2=49 / 4
$
Example 3: If a circle passing through the point $(-1,0)$ touches the $y$-axis at $(0,2)$, then the length of the chord of the circle along the $x$-axis is :
Solution:
As learnt in the concept
Circle touching $y$-axis and having radius $r$ -
$
x^2+y^2+2 g x \pm 2 r y+g^2=0
$
- wherein
Where g is a variable parameter.
Equation of a circle -
$
(x-h)^2+(y-k)^2=r^2
$
- wherein
Circle with centre $(h, k)$ and radius $r$.
If the centre is $(h, 2)$ then
$
\text { radius }=|\mathrm{h}|
$
equation of a circle is
$
(x-h)^2+(y-2)^2=h^2
$
and it passes through point $(-1,0)$
putting values, we get
$
h=\frac{-5}{2}
$
So centre is
$
\left(\frac{-5}{2}, 2\right)
$
equation $\left(x+\frac{5}{2}\right)^2+(y-2)^2=\left(\frac{5}{2}\right)^2$
$A B$ is a chord along the $x$-axis
$
\mathrm{AB}=2(\mathrm{AM})=2 \sqrt{\frac{25}{4}-4}=3
$
Example 4: $\mathbf{A}$ variable circle passes through the fixed point $A(p, q)$ and touches the $\mathbf{x}$-axis. The locus of the other end of the diameter through $A$ is
Solution:
Let the other diametric end be $\mathrm{P}(\mathrm{h}, \mathrm{k})$
So centre is $\left(\frac{p+h}{2}, \frac{q+R}{2}\right)$
Radius $=\sqrt{\left(\frac{h-p}{2}\right)^2+\left(\frac{k-q}{2}\right)^2}$
For a circle touching the $x$-axis, radius $=\left(\frac{q+k}{2}\right)$
So $\left(\frac{h-p}{2}\right)^2+\left(\frac{k-q}{2}\right)^2=\left(\frac{k+q}{2}\right)^2$
we get $(h-p)^2=4 \mathrm{~kg}$
i.e. $(x-p)^2=4 q y$. a parabola
Example 5: The lines $2 x-3 y=5$ and $3 x-4 y=7$ are diameters of a circle having an area of 154 sq. units. Then the equation of the circle is
Solution:
The Centre is a point of intersection of
$
\begin{aligned}
& 2 x-3 y=5 \text { and } 3 x-4 y=7 \\
& \text { i.e. } x=1, y=-1 \\
& \text { also } \pi r^2=154 \\
& \frac{22}{7} \times r^2=154 \\
& \Rightarrow r^2=49 \\
& \Rightarrow r=7
\end{aligned}
$
equation of the circle is
$
\begin{aligned}
& (x-1)^2+(y+1)^2=7^2 \\
& x^2+y^2-2 x+2 y=47
\end{aligned}
$
Frequently Asked Questions (FAQs)