Chord of Contact: Definition, Theorem, Equation, Formula

The concept of a chord bisected at a given point helps in deriving important properties and equations related to circles. Specifically, the equation of such a chord allows for determining its precise location and length, given its midpoint and the circle’s equation. Understanding how to derive and utilize the equation of a chord bisected at a given point is essential for solving complex geometric problems involving circles. It forms the basis for various applications, from calculating the length of a chord to analyzing its position relative to the centre of the circle.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

This Story also Contains

1. Equation of Chord Bisected at a Given Point
2. Solved Examples Based on Equation of Chord Bisected at a Given Point
3. Summary

Equation of Chord Bisected at a Given Point

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)

A chord of a circle is a line segment with both endpoints on the circle. An interesting geometric property of chords is that they can be uniquely defined by their midpoint when they are bisected at that point.

The equation of the chord (AB) of the circle S = 0 whose midpoint is M(x₁,y₁) is

$\mathrm{xx}_1+\mathrm{yy}_1-\mathbf{a}^2=\mathrm{x}_1^2+\mathbf{y}_1^2-\mathbf{a}^2 \quad$ or $\quad \mathbf{T}=\mathbf{S}_1$

coordinate of point O is $(0,0)$ and Point M is $\left(x_1, y_1\right)$
slope of $O M=\frac{0-y_1}{0-\mathrm{x}_1}=\frac{\mathrm{y}_1}{\mathrm{x}_1}$
$\therefore$ slope of $\mathrm{AB}=-\frac{\mathrm{x}_1}{\mathrm{y}_1}$
the, equation of $A B$ is $y-y_1=-\frac{x_1}{y_1}\left(x-x_1\right)$
or

$\mathrm{yy}_1-\mathrm{y}_1^2=\mathrm{xx}_1+\mathrm{x}_1^2$

or

$\mathrm{xx}_1+\mathrm{yy}_1-\mathrm{a}^2=\mathrm{x}_1^2+\mathrm{y}_1^2-\mathrm{a}^2 \quad \text { or } \quad \mathbf{T}=\mathbf{S}_1$

Note:

Same result $T=S_1$ can also be applied to get a chord bisected at a given point for any circle of type

$x^2+y^2+2 g x+2 f y+c=0$ as well. as well.

Recommended Video Based on Equation of Chord Bisected at a Given Point

Solved Examples Based on Equation of Chord Bisected at a Given Point

Example 1: Let the tangents at two points A and B on the circle $\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+3=0$ meet at origin $\mathrm{O}(0,0)$. Then the area of the triangle OAB is:
1) $\frac{3 \sqrt{3}}{2}$
2) $\frac{3 \sqrt{3}}{4}$
3) $\frac{3}{2 \sqrt{3}}$
4) $\frac{3}{4 \sqrt{3}}$

Solution

$\begin{aligned}
& \mathrm{r}=\sqrt{4-3}=1 \\
& \mathrm{OA}=\sqrt{\mathrm{s}_1}=\sqrt{3} \\
& \therefore \tan \theta=\frac{1}{\sqrt{3}}
\end{aligned}$

$\begin{aligned}
& \text { Area of } \mathrm{OAB}=2 \times\left(\frac{1}{2} \times \sqrt{3} \cos \theta \cdot \sqrt{3} \sin \theta\right) \\
& =3 \sin \theta \cdot \cos \theta \\
& =3 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} \\
& =\frac{3 \sqrt{3}}{4}
\end{aligned}$

$\therefore$ Option (B)

Example 2: Let AB be a chord of length 12 of the circle $(x-2)^2+(y+1)^2=\frac{169}{4}$. If tangents drawn to the circle at points A and B intersect at the point P , then five times the distance of point P from chord AB is equal to

1) 72

2) 30

3) 34

4) 65

Solution

$\begin{aligned}
& \sin \theta=\frac{\mathrm{BE}}{\mathrm{BC}}=\frac{6}{\frac{13}{2}}=\frac{12}{13} \\
& \text { Now } \frac{\mathrm{BE}}{\rho \mathrm{E}}=\tan (90-\theta) \\
& \Rightarrow \frac{6}{\mathrm{PE}}=\cot \theta \\
& \Rightarrow \mathrm{PE}=6 \tan \theta=6 \cdot \frac{12}{5}=\frac{72}{5} \\
& \Rightarrow 5 \mathrm{PE}=72 \\
& \text { Ans }: 72
\end{aligned}
$
Example 3: Let B be the centre of the circle $x^2+y^2-2 x+4 y+1=0$. Let the tangents at two points P and Q on the circle intersect at the point $\mathrm{A}(3,1)$. Then $\left.8 \cdot\left(\frac{\text { area }}{\text { area } \triangle \mathrm{APQ}}\right) \mathrm{BPQ}\right)$ is equal to $\qquad$

1) 18

2) 34

3) 6

4) 7

Solution

$B=(1,-2), r=\sqrt{1+4-1}=2$

$
\begin{aligned}
& \triangle A P Q=\frac{1}{2} L^2 \sin \theta \text { where L=length of tangent } \\
& =\sqrt{S_1} \\
& =\sqrt{9+1-6+4+1} \\
& =3 \\
& \Rightarrow \triangle A P Q=\frac{1}{2} \times 9 \sin \theta \\
& \triangle B P Q=\frac{1}{2} R^2 \sin (\pi-\theta)=\frac{1}{2} \times 4 \sin \theta \\
& 8 \times \frac{\Delta A P Q}{\Delta B P Q}=8 \times \frac{9}{4}=18
\end{aligned}
$

Example 4: The equation of the chord of contact of the origin w.r.t. circle $x^2+y^2-2 x-4 y-4=0$ is
1) $x+2 y+4=0$
2) $x+2 y-4=0$
3) no chord of contact exist
4) none of these

Solution
Fince $S(0,0)=-4<0$
$\therefore \quad(0,0)$ is inside the circle. So no chord of contact exists.
Hence, the answer is the option (3).

Example 5: Equation of chord AB of circle $\mathrm{x}^2+\mathrm{y}^2=2$ passing through $\mathrm{P}(2,2)$ such that $\mathrm{PB} / \mathrm{PA}=3$, is given by:
1) $x=3 y$
2) $x=y$
3) $y-2=\sqrt{3}(x-2)$
4) none of these

Solution
Key concept: Using the concept of the parametric equation of any line. PB and PA. Let the roots be $\mathrm{r}_1$ and $\mathrm{r}_2$.
Then $\frac{\mathrm{PB}}{\mathrm{PA}}=\frac{\mathrm{r}_2}{\mathrm{r}_1}$, now if $\mathrm{r}_1=\alpha, \mathrm{r}_2=3 \alpha$.
then $4 \alpha=-4(\sin \theta+\cos \theta), 3 \alpha^2=6 \Rightarrow \sin 2 \theta=1 \Rightarrow \theta=\pi / 4$
So, the required chord will be $\mathrm{y}-2=1(\mathrm{x}-2) \Rightarrow \mathrm{y}=\mathrm{x}$

Alternative solution:
Key concept: Using the basic property of a circle.

$
\mathrm{PA} \cdot \mathrm{PB}=\mathrm{PT}^2=2^2+2^2-2=6
$
$
\frac{\mathrm{PB}}{\mathrm{PA}}=3
$

From (1) and (2), we have

$
\mathrm{PA}=\sqrt{2}, \mathrm{~PB}=3 \sqrt{2}
$

$\Rightarrow \mathrm{AB}=2 \sqrt{2}$. Now the diameter of the circle is $2 \sqrt{2}$ (as the radius is $\sqrt{2}$ )
Hence line passes through the centre

$
\Rightarrow \mathrm{y}=\mathrm{x}
$

Hence, the answer is the option (2)

Summary

The equation of a chord bisected at a given point provides a powerful tool in geometry, offering insights into various geometric properties and relationships involving circles. The derivation not only strengthens the understanding of circle properties but also facilitates the solving of complex geometric problems. Whether in theoretical explorations or practical applications, the concept holds significant value in the study of geometry.