Pair of Tangent: Definition, Equation and Formula

In geometry, a circle is a fundamental shape defined as the locus of all points that are equidistant from a fixed point, known as the center. One interesting property of circles is the concept of tangents — lines that touch the circle at exactly one point. A special case arises when dealing with tangents drawn from a point outside the circle. The study of these tangents provides insights into various geometric properties and relationships.

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This Story also Contains

1. Pair of Tangent
2. Solved Examples Based on Pair of Tangent
3. Summary

Pair of Tangent

A tangent to a circle is a straight line that touches the circle at exactly one point. This point of contact is known as the point of tangency. The tangent is perpendicular to the radius of the circle at the point of contact. For a given point P(x1,y1) outside a circle, there are typically two distinct tangents that can be drawn to the circle. These tangents are crucial in various geometric constructions and proofs.

If the line L touches the circle, then Equation (iii) will have two equal real roots

So, Discriminant of equation (iii) = 0

$\begin{aligned} & \mathrm{B}^2-4 \mathrm{AC}=0 \\ & 4 \mathrm{~m}^2 \mathrm{c}^2-4\left(1+\mathrm{m}^2\right)\left(\mathrm{c}^2-\mathrm{a}^2\right)=0 \\ & \mathrm{a}^2=\frac{\mathrm{c}^2}{1+\mathrm{m}^2} \\ & c^2=a^2\left(1+m^2\right)\end{aligned}$

In this case, the line is a tangent to the circle

This is also the condition of tangency to the circle.

Equation of the Tangent in Point Form

Point Form

The equation of the tangent to a circle $x^2+y^2+2 g x+2 f y+c=0$ at the point $P\left(x_1 \cdot y_1\right)$ is $x_1+y y_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0$
Proof:
$\mathrm{C}(-\mathrm{g},-\mathrm{f})$ is the centre of the circle
As point $\mathrm{P}\left(x_1, y_1\right)$ lies on the circle.
$\therefore \quad$ Slope of $\mathrm{CP}=\frac{\mathrm{y}_1-(-\mathrm{f})}{\mathrm{x}_1-(-\mathrm{g})}=\frac{\mathrm{y}_1+\mathrm{f}}{\mathrm{x}_1+\mathrm{g}}$
Here, PT is the perpendicular to CP.
Thus, $\quad$ slope of $\mathrm{PT}=-\left(\frac{\mathrm{x}_1+\mathrm{g}}{\mathrm{y}_1+\mathrm{f}}\right)$
Hence, the equation of the tangent at $\mathrm{P}\left(x_1, y_1\right)$ is

$
\begin{gathered}
\left(y-y_1\right)=-\left(\frac{x_1+g}{y_1+f}\right)\left(x-x_1\right) \\
\Rightarrow \quad\left(y-y_1\right)\left(y_1+f\right)+\left(x_1+g\right)\left(x-x_1\right)=0 \\
\Rightarrow \quad x_1+y_1+g x+f y=x_1^2+y_1^2+g x_1+\mathrm{yy}_1
\end{gathered}
$

now add $\mathrm{gx}_1+\mathrm{fy}_1+\mathrm{c}$ both side, we get

$
\Rightarrow \quad \mathrm{xx}_1+\mathrm{yy}_1+\mathrm{g}\left(\mathrm{x}+\mathrm{x}_1\right)+\mathrm{f}\left(\mathrm{y}+\mathrm{y}_1\right)+\mathrm{c}=\mathrm{x}_1^2+\mathrm{y}_1^2+2 g \mathrm{x}_1+2 \mathrm{yy}_1+\mathrm{c}
$

i.e. $\quad x_1+y_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0$
(As, point $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ lies on the circle so, $\mathrm{x}_1^2+\mathrm{y}_1^2+2 g \mathrm{x}_1+2 f \mathrm{y}_1+\mathrm{c}=0$ )

NOTE:

In order to find out the equation of a tangent to any 2nd-degree curve, the following points must be kept in mind:

$x_1^2$ is replaced by $x x_1$
$y^2$ is replaced by $y y_1$
$x y$ is replaced by $\frac{x y_1+x_1 y}{2}$ $x$ is replaced by $\frac{x+x_1}{2}$ $y$ is replaced by $\frac{y+y_1}{2}$
and c vill remain $c$.
This method is applicable only for a 2nd-degree conic.
Pair of Tangent:
The combined equation of the pair of tangents drawn from $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ to the circle $\mathrm{S}: \mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ is

$
\left(x^2+y^2-a^2\right)\left(x_1^2+y_1^2-a^2\right)=\left(x x_1+y y_1-a^2\right)^2 \quad \text { or } \quad S S_1=T^2
$

Where,

$
\begin{aligned}
& S \equiv x^2+y^2-a^2 \\
& S_1 \equiv x_1^2+y_1^2-a^2 \\
& T \equiv x x_1+y y_1-a^2
\end{aligned}
$

The combined equation of a pair of tangents drawn from $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ to a circle $x^2+y^2+2 g x+2 f y+c=0$ is

$
\begin{aligned}
& \quad S S_1=T^2 \\
& \text { where } \\
& \qquad S \equiv x^2+y^2+2 g x+2 f y+c \\
& S_1 \equiv x_1^2+y_1^2+2 g x_1+2 f y_1+c \\
& T \equiv x x_1+y y_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c
\end{aligned}
$

Recommended Video Based on Pair of Tangent

Solved Examples Based on Pair of Tangent

Example 1: The area of the triangle formed by the pair of tangents from $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ to the circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ and the chord joining their points of contact with the circle is $\mathrm{k}\left(\mathrm{x}_1^2+\mathrm{y}_1^2-\mathrm{a}^2\right)^{3 / 2}\left(\mathrm{x}_1^2+y_1^2\right)^{-1}$ sq. units where $\mathrm{k}=$
1) $2 a$
2) $\frac{a}{2}$
3) $a$
4) 2

Solution
Let $R\left(x_1, y_1\right)$ be the point from which are drawn the tangents $R P$ and $R Q$ to the circle $2 k\left(x_1^2+y_1^2-a^2\right)^{3 / 2}\left(x_1^2+y_1^2\right)^{-1}$ sq. units.
Let $O R$ meet $P Q$ at $L$, where $O(0,0)$ is the centre of the circle.

$
\begin{aligned}
& \text { Then } \triangle \mathrm{RPQ}=2 \Delta \mathrm{RLP}=2[\Delta \mathrm{OPR}-\Delta \mathrm{OLP}] \\
& \qquad \begin{aligned}
\Delta \mathrm{OPR}=1 / 2 \mathrm{OP} \cdot \mathrm{PR}=1 / 2 \mathrm{a} \cdot \sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2-\mathrm{a}^2} \\
\text { but } \Delta \mathrm{OLP}=1 / 2 \mathrm{OL} \cdot \mathrm{LP}=1 / 2(\mathrm{OP} \cos \theta)(\mathrm{OP} \sin \theta) \text { where } \angle \mathrm{POR}=\theta
\end{aligned}
\end{aligned}
$

$\Delta \mathrm{OPR}, \sin \theta=\frac{\mathrm{PR}}{\mathrm{OR}}=\frac{\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2-\mathrm{a}^2}}{\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2}}$ and$\cos \theta=\frac{\mathrm{OP}}{\mathrm{OR}}=\frac{\mathrm{a}}{\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2}}$

$
\begin{array}{ll}
& \frac{\sqrt{x_1^2+y_1^2-a^2}}{2\left(x_1^2+y_1^2\right)} a^3 \\
\text { hence } \Delta \mathrm{LOP}=1 / 2 \mathrm{a}^2 \cos \theta \sin \theta=\quad & \text { and } \triangle \mathrm{RPQ}=2\left[\frac{1}{2} a \sqrt{x_1^2+y_1^2-a^2}-\frac{1}{2} a^3 \frac{\sqrt{x_1^2+y_1^2-a^2}}{x_1^2+y_1^2}\right] \\
=a \sqrt{x_1^2+y_1^2-a^2}\left[1-\frac{a^2}{x_1^2+y_1^2}\right]=a \sqrt{x_1^2+y_1^2-a^2} \frac{\left(x_1^2+y_1^2-a^2\right)}{x_1^2+y_1^2}=\frac{a\left(x_1^2+y_1^2-a^2\right)^{3 / 2}}{x_1^2+y_1^2}
\end{array}
$

Example 2: The angle between tangents from the origin to the circle $(x-7)^2+(y+1)^2=25$ is
1) $\pi / 3$
2) $\pi / 2$
3) $\pi / 6$
4) 0

Solution

$
\begin{aligned}
& \sin \theta=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4} \\
& \text { Angle between tangents }=\frac{\pi}{2} \text {. }
\end{aligned}
$

Hence (C) is the correct answer.

Example 3: The tangents to $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ having inclinations $\alpha$ and $\beta$ intersect at P . If $\cot \alpha+\cot \beta=0$, then the locus of P is
1) $x+y=0$
2) $x-y=0$
3) $x y=0$
4) None of these.

Solution
Let the coordinates of P be $(\mathrm{h}, \mathrm{k})$. Let the equation of a tangent from $\mathrm{P}(\mathrm{h}, \mathrm{k})$ to the circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ be

$
\mathrm{y}=\mathrm{mx}+\mathrm{a} \sqrt{1+\mathrm{m}^2}
$Since, $P(h, k)$ lies on $y=m x+a \sqrt{1+m^2}$, therefore,

$
\begin{aligned}
& \mathrm{k}=\mathrm{mh}+\mathrm{a} \sqrt{1+\mathrm{m}^2} \mathrm{P}(k-m h)^2=a \sqrt{(1+m)^2} \\
& \mathrm{pm}^2\left(\mathrm{k}^2-\mathrm{a}^2\right)-2 m k h+\mathrm{h}^2-\mathrm{a}^2=0
\end{aligned}
$

The is a quadratic in m . Let the two roots be $\mathrm{m}_1$ and $\mathrm{m}_2$, then

$
\mathrm{m}_1+\mathrm{m}_2=\frac{2 \mathrm{hk}}{\mathrm{k}^2-\mathrm{a}^2}
$

But $\tan \alpha=\mathrm{m}_1, \tan \beta=\mathrm{m}_2$ and it's given that

$
\cot \alpha+\cot \beta=0
$
$
\frac{1}{m_1}+\frac{1}{m_2}=0 \mathrm{pm}_1+m_2=0 \quad p \frac{2 h k}{k^2-a^2}=0
$

$\Rightarrow \quad \mathrm{hk}=0$. Hence, the locus of $(\mathrm{h}, \mathrm{k})$ is $\mathrm{xy}=0$.
Hence, the answer is the option (3).

Example 4: The slope of a common tangent to the ellipse $\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ and a concentric circle of radius $r$ is (where $b<r<a$ )
1) $\tan ^{-1} \sqrt{\frac{r^2-b^2}{a^2-r^2}}$
2) $\sqrt{\frac{r^2-b^2}{a^2-r^2}}$
3) $\left(\frac{r^2-b^2}{a^2-r^2}\right)$
4) $\sqrt{\frac{a^2-r^2}{r^2-b^2}}$

Solution
$y=m x+\sqrt{a^2 m^2+b^2}$ is tangent to the ellipse.
Equation of concentric circle be $x^2+y^2=r^2$
$\mathrm{y}=\mathrm{mx} \pm \mathrm{r} \sqrt{1+\mathrm{m}^2}$ is tangent to the circle.

.Equation of concentric circle be $x^2+y^2=r^2$
$y=m x \pm r \sqrt{1+m^2}$ is tangent to the circle.

$
\begin{array}{cc}
\therefore \quad & r \sqrt{1+m^2}=\sqrt{a^2 m^2+b^2} \\
& \left(1+m^2\right) r^2=a^2 m^2+b^2 \\
& m^2\left(r^2-a^2\right)=b^2-r^2 \\
& m=\sqrt{\frac{r^2-b^2}{a^2-r^2}}
\end{array}
$

Hence, the answer is the option (2).

Example 5: From a point T , two mutually perpendicular tangents TA and TB are drawn to the parabola $\mathrm{y}^2=4 \mathrm{ax}$. If the minimum area of the circle having AB as diameter is k , then $\frac{\mathrm{k}}{\pi \mathrm{a}^2}$ is
1) 4
2) 1
3) 2
4) 8

Solution
Since tangents T A and T B are mutually perpendicular, circle drawn on A B as diameter passes through T. Hence, A B will be focal chord of the parabola. If A $\equiv\left(\right.$ at $\left.{ }^2, 2 a t\right)$, then

$
B \equiv\left(\frac{a}{t^2},-\frac{2 a}{t}\right)
$

Length of focal chord will be minimum when $t= \pm 1 \Rightarrow A \equiv(a, 2 a), B \equiv(a,-2 a) \Rightarrow A B$ will be the latus rectum. Hence, area of the circle is $\pi(2 \mathrm{a})^2=4 \pi \mathrm{a}^2$.
Hence, the answer is the option (1).

Summary

The concept of tangents to a circle, particularly the combined equation of the pair of tangents drawn from an external point, is a vital aspect of geometry. By deriving and applying this equation, one can solve a variety of geometric problems and gain deeper insights into the properties of circles and tangents.