Locus of Mid the Point of a Chord of the Circle

A circle is one of the most fundamental geometric shapes, consisting of all points in a plane that is equidistant from a fixed point called the centre of a circle. It is a very basic shape that is constantly used in mathematics. The main applications of the circle are in geometry, engineering for designing circular instruments, physics, and technology.

In this article, we will cover the concept of the locus of the midpoint of the chord 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 three questions have been asked on this concept, including three in 2023,

Locus and its Equation

When a point moves in a plane under certain geometrical conditions, then the point traces out a path, This path of the moving point is known as the locus of this point.

For example, let a point O(0,0) be a fixed point (i.e. origin) and a variable point P (x, y) is in the same plane. If point P moves in such a way that the distance OP is constant r, then point P traces out a circle whose centre is O(0, 0) and radius is r.

Steps to Finding the Equation of Locus

1. Consider the point (h, k) whose locus is to be found.

2. Express the given condition as an equation in terms of the known quantities and unknown parameters.

3. Eliminate the parameters so that the resultant equation consists only of locus coordinates h, k, and known quantities.

4. Now, replace the locus coordinate (h, k) with (x, y) in the resultant equation.

Locus of Mid Point of the Chord of the Circle

A circle with radius r is centred at the point (h, k), and AB is its chord. Let M (x₁, y₁) be the midpoint of the chord AB.

From the figure,

$
\begin{aligned}
& \cos \theta=\frac{\mathrm{CM}}{r}=\frac{\sqrt{\left(\mathrm{x}_1-\mathrm{h}\right)^2+\left(\mathrm{y}_1-\mathrm{k}\right)^2}}{r} \\
& \Rightarrow 1-\sin ^2 \theta=\frac{\left(\mathrm{x}_1-\mathrm{h}\right)^2+\left(\mathrm{y}_1-\mathrm{k}\right)^2}{r^2} \\
& \Rightarrow-\sin ^2 \theta=\frac{\left(\mathrm{x}_1-\mathrm{h}\right)^2+\left(\mathrm{y}_1-\mathrm{k}\right)^2-\mathrm{r}^2}{r^2}
\end{aligned}
$

$\therefore$ Required equation of locus is

$
\Rightarrow \quad \frac{(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2-\mathrm{r}^2}{r^2}=-\sin ^2 \theta
$

Recommended Video Based on Locus of Mid Point of the Chord of the Circles

Solved Examples Based on Locus of Mid Point of the Chord of the Circles

Example 1: Find the condition on $a, b, c$ such that two chords of the circle $x^2+y^2-2 a x-2 b y+a^2+b^2-c^2=0$ passing through the point $(a, b+c)$ are bisected by the line $y=x$.
1) $4 \mathrm{a}^2+4 \mathrm{~b}^2-4 \mathrm{c}^2-8 \mathrm{ab}+\mathrm{bc}-\mathrm{ac}<0$
2) $4 a^2+4 b^2-8 a b+4 b c-4 a c-c^2<0$
3) $4 a^2+2 b^2-4 a b+4 b c-a c+c^2>0$
4) $4 a^2+4 b^2-8 a b-c^2>0$

Solution:
Chords are bisected on the line $\mathrm{y}=\mathrm{x}$. Let $\left(\mathrm{x}_1, \mathrm{x}_1\right)$ be the midpoint of the chord then the equation of the chord is $\mathrm{T}=\mathrm{S}_1$

$
\begin{aligned}
& \therefore \quad \mathrm{xx}_1+\mathrm{yx}_1-\mathrm{a}\left(\mathrm{x}+\mathrm{x}_1\right)-\mathrm{b}\left(\mathrm{y}+\mathrm{x}_1\right)+\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2 \\
& =\mathrm{x}_1^2+\mathrm{x}_1^2-2 \mathrm{ax}_1-2 \mathrm{bx}_1+\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2 \\
& \Rightarrow \quad\left(\mathrm{x}_1-\mathrm{a}\right) \mathrm{x}+\left(\mathrm{x}_1-\mathrm{b}\right) \mathrm{y}+\mathrm{ax}_1+\mathrm{bx}_1-2 \mathrm{x}_1^2=0
\end{aligned}
$
This chord passes through $(\mathrm{a}, \mathrm{b}+\mathrm{c})$

$
\begin{aligned}
& \Rightarrow \quad\left(\mathrm{x}_1-\mathrm{a}\right) \mathrm{a}+\left(\mathrm{x}_1-\mathrm{b}\right)(\mathrm{b}+\mathrm{c})+\mathrm{ax}_1+\mathrm{bx}_1-2 \mathrm{x}_1^2=0 \\
& \Rightarrow \quad 2 \mathrm{x}_1^2-(2 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}) \mathrm{x}_1+\mathrm{a}^2+\mathrm{b}^2+\mathrm{bc}=0
\end{aligned}
$

which is quadratic in $\mathrm{x}_1$. Since it is given that two chords are bisected on the line $\mathrm{y}=\mathrm{x}$, then $\mathrm{x}_1$ must have two distinct real roots,

$
\begin{aligned}
& \therefore \mathrm{B}^2-4 A C>0 \\
& \Rightarrow(2 a+2 b+c)^2-4 \cdot 2 \cdot\left(a^2+b^2+b c\right)>0 \\
& \Rightarrow 4 a^2+4 b^2+c^2+8 a b+4 b c+4 a c-8 a^2-8 b^2-8 b c>0 \\
& \Rightarrow 4 a^2+4 b^2-8 a b+4 b c-4 a c-c^2<0
\end{aligned}
$
Hence the condition on $\mathrm{a}, \mathrm{b}, \mathrm{c}$ is

$
4 a^2+4 b^2-c^2-8 a b+4 b c-4 a c<0
$
Hence, the answer is the option 2.

Example 2: The locus of the point of intersection of the tangents to the circle $x=r \cos \theta, y=r \sin \theta$ at points whose parametric angles are differ by $\pi / 3$, is
Solution:
Let one of the points on the circle be $\mathrm{P}(\mathrm{r} \cos \theta, \mathrm{r} \sin \theta)$.
Then the other point is $\mathrm{Q}\{\mathrm{r} \cos (\theta+\pi / 3), \mathrm{r} \sin (\theta+\pi / 3)\}$.
$\therefore$ Equation of the tangent at P is $\mathrm{x} \cos \theta+\mathrm{y} \sin \theta=\mathrm{r}$,
and the equation of the tangent at

$
\mathrm{Q} \text { is } \mathrm{x} \cos \left(\theta+\frac{\pi}{3}\right)+\mathrm{y} \sin \left(\theta+\frac{\pi}{3}\right)=\mathrm{r}
$
$
\begin{aligned}
& \Rightarrow \quad \mathrm{x}\left[\cos \theta\left(\frac{1}{2}\right)-\sin \theta\left(\frac{\sqrt{3}}{2}\right)\right]+\mathrm{y}\left[\sin \theta\left(\frac{1}{2}\right)+\cos \theta\left(\frac{\sqrt{3}}{2}\right)\right]=\mathrm{r} \\
& \Rightarrow \quad \frac{1}{2}(\mathrm{x} \cos \theta+\mathrm{y} \sin \theta)-\frac{\sqrt{3}}{2}(\mathrm{x} \sin \theta-\mathrm{y} \cos \theta)=\mathrm{r} \\
& \Rightarrow \quad \frac{\mathrm{r}}{2}-\frac{\sqrt{3}}{2}(\mathrm{x} \sin \theta-\mathrm{y} \cos \theta)=\mathrm{r} \quad \Rightarrow \quad(\mathrm{x} \sin \theta-\mathrm{y} \cos \theta)=-\frac{\mathrm{r}}{\sqrt{3}}
\end{aligned}
$
The locus of the point of intersection of this tangent can now be obtained by eliminating $\theta$
$\therefore$ Squaring and adding (i) and (ii)

$
\begin{aligned}
& \Rightarrow \quad(\mathrm{x} \cos \theta+\mathrm{y} \sin \theta)^2+(\mathrm{x} \sin \theta-\mathrm{y} \cos \theta)^2=\mathrm{r}^2+\left(-\frac{\mathrm{r}}{\sqrt{3}}\right)^2 \\
& \Rightarrow \quad 3\left(\mathrm{x}^2+\mathrm{y}^2\right)=4 \mathrm{r}^2
\end{aligned}
$
Example 3: The circle $\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-4 \mathrm{y}+4=0$ is inscribed in a triangle which has two of its sides along the coordinate axes. The locus of the circum-center of the triangle is $\mathrm{x}+\mathrm{y}-\mathrm{xy}+\mathrm{k}\left(\mathrm{x}^2+\mathrm{y}^2\right)^{1 / 2}=0$ Find k .
Solution:
Let $O A B$ be the triangle that has two sides $O A$ and $O B$ along the axes. Let the equation of the third side $A B$ be $\frac{x}{a}+\frac{y}{b}=1$
$\therefore$ coordinate of vertices are $\mathrm{O}(0,0), \mathrm{A}(\mathrm{a}, 0)$ and $\mathrm{B}(0, \mathrm{~b})$
The circle inscribed in the $\triangle O A B$ is

$
x^2+y^2-4 x-4 y+4=0
$

whose center is $(2,2)$ and radius $=2$.
Since $(2,2)$ is the incentre of $\triangle O A B$, we have

$
\begin{aligned}
& \frac{0 \cdot \sqrt{\mathrm{a}^2+\mathrm{b}^2}+\mathrm{a} \cdot \mathrm{b}+0 \cdot \mathrm{a}}{\mathrm{a}+\mathrm{b}+\sqrt{\mathrm{a}^2+\mathrm{b}^2}}=2 \\
& \frac{0 \cdot \sqrt{\mathrm{a}^2+\mathrm{b}^2}+\mathrm{b} \cdot \mathrm{a}+0 \cdot \mathrm{b}}{\mathrm{a}+\mathrm{b}+\sqrt{\mathrm{a}^2+\mathrm{b}^2}}=2 \\
& \therefore \mathrm{ab}=2\left[\mathrm{a}+\mathrm{b}+\sqrt{\mathrm{a}^2+\mathrm{b}^2}\right]
\end{aligned}
$
If $P(\alpha, \beta)$ is the circum-center of $\triangle O A B$, then
$\therefore$ locus of $P(\alpha, \beta)$ is

$
x+y-x y+\sqrt{x^2+y^2}=0
$
But it is given that locus of circumcentre $P(\alpha, \beta)$ is

$
x+y-x y+k \sqrt{x^2+y^2}=0
$
Comparing (2) and (3), we get $\mathrm{k}=1$.

Example 4: The locus of the midpoint of the chord of the circle $x^2+y^2=9$ such that the segment intercepted by the chord on the curve $y^2-4 x-4 y=0$ subtends the right angle at the origin is Solution:
equation of chord whose midpoint is ( $h, k$ ) is $h \mathrm{x}+\mathrm{ky}=\mathrm{h}^2+\mathrm{k}_{\text {making homogenous eq with parabola }}^2$

$
y^2-4(x+y) \frac{(h x+k y)}{h^2+k^2}=0
$
Coeff $x^2+$ coeff $y^2=0 \quad h^2+k^2=4(h+k)$

$
\begin{aligned}
& 4 \alpha \beta=2\left[2 \alpha+2 \beta+2 \sqrt{\alpha^2+\beta^2}\right]
\end{aligned}
$

Summary

The circles are foundational shapes with unique properties and applications in various mathematics, science, and engineering fields. Understanding the properties, equations, and applications of circles is essential for solving geometric problems, designing objects, and analyzing natural phenomena.