Tangent to a Circle

In geometry, the concept of a tangent to a circle is essential for understanding geometric relationships and solving various problems involving circles. The tangent to a circle at a given point is a straight line that touches the circle at exactly one point without crossing it. This tangent line is perpendicular to the radius of the circle at the point of tangency. The equation of the tangent to a circle can be derived and expressed in several forms depending on the given information.

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Equation of the tangent to a Circle

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 center (O) of the circle and the constant distance is called its radius ($r$)

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 tangent to the circle

This is also the condition of tangency to the circle.

Equation of the Tangent in 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 $P\left(x_1, y_1\right)$ is $x_1+y_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0$

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

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

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

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

Note:

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

$x^2$ 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 $\frac{x+x_1^2}{2}$
$y$ is replaced by $\frac{y+y_1}{2}$

and c will remain c.

This method is applicable only for a 2nd degree conic.

Equation of Tangent of Circle in Parametric Form

The equation of the tangent at the point $(a \cos \theta, \mathrm{a} \sin \theta)$ to a circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ is $\mathrm{x} \cos \theta+\mathrm{y} \sin \theta=\mathbf{a}$
Proof:
If $S=x^2+y^2-a^2=0$ is the circle, then the tangent at $\left(x_1, y_1\right)$ is $T_1=$ $\mathrm{xx}_1+\mathrm{yy}_1-\mathrm{a}^2=0$
put, $\mathrm{x}_1=\mathrm{a} \cos \theta, \quad \mathrm{y}_1=\mathrm{a} \sin \theta$
we get, $\quad \mathrm{x} \cos \theta+\mathrm{y} \sin \theta=\mathrm{a}$

Equation of the Tangent in Slope Form

The equation of the tangent to a circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ having slope m is $\mathrm{y}=\mathrm{mx} \pm \mathrm{a} \sqrt{\left(\mathbf{1 + \mathbf { m } ^ { 2 } )}\right.}$, and point of tangency is $\left( \pm \frac{a m}{\sqrt{\left(1+m^2\right)}}, \mp \frac{a}{\sqrt{\left(1+m^2\right)}}\right)$.
Let $y=m x+c$ be a tangent to the circle $x^2+y^2=a^2$.
$\therefore \quad$ Length of perpendicular from centre of circle $(0,0)$
on $(y=m x+c)=$ radius of circle

$\therefore \quad \frac{|c|}{\sqrt{1+\mathrm{m}^2}}=\mathrm{a} \Rightarrow \mathrm{c}= \pm \mathrm{a} \sqrt{1+\mathrm{m}^2}$

substituting this value of $c$ in $y=m x+c$, we get $\mathbf{y}=\mathbf{m x} \pm \mathbf{a} \sqrt{\left(\mathbf{1}+\mathbf{m}^2\right)}$
which are the required equations of tangents.

Corollary: It also follows that $y=m x+c$ is tangent to $x^2+y^2=a^2$ if $c^2=a^2\left(1+m^2\right)$ which is the condition of tangency.

Point of Contact:

Solving $x^2+y^2=a^2$ and $y=m x \pm a \sqrt{1+m^2}$, simultaneously we get,

$\begin{aligned}
& x= \pm \frac{a m}{\sqrt{\left(1+m^2\right)}} \\
& y=\mp \frac{a}{\sqrt{\left(1+m^2\right)}}
\end{aligned}$

Thus, the coordinates of the points of contact are

$\left( \pm \frac{a m}{\sqrt{\left(1+m^2\right)}}, \mp \frac{a}{\sqrt{\left(1+m^2\right)}}\right)$

NOTE:
Equation of tangent of the circle $(x-h)^2+(y-k)^2=a^2$ in terms of slope is $(y-k)=m(x-h) \pm a \sqrt{\left(1+m^2\right)}$

Solved Examples Based on Equation of the tangent to a Circle:

Example 1: Let the tangent to the circle $x^2+y^2=25$ at the point $R(3,4)$ meet $x$-axis and $y$-axis at points P and Q , respectively. If $r$ is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ , then $r^2$ is equal to :
1) $\frac{529}{64}$
2) $\frac{585}{66}$
3) $\frac{125}{72}$
4) $\frac{625}{72}$

Solution

Tangent to the circle x² + y² = 25 at R(3, 4) is 3x + 4y = 25

$\begin{aligned} & I \equiv\left(\frac{\frac{625}{12}}{\frac{25}{4}+\frac{25}{3}+\frac{125}{12}}, \frac{\frac{625}{12}}{\frac{25}{4}+\frac{25}{3}+\frac{125}{12}}\right) \\ & \therefore I \equiv\left(\frac{625}{75+100+125}, \frac{625}{75+100+125}\right) \equiv\left(\frac{25}{12}, \frac{25}{12}\right) \\ & \therefore r^2=\left(\frac{25}{12}\right)^2+\left(\frac{25}{12}\right)^2=\frac{625}{72}\end{aligned}$

Example 2: Find the equation to the tangent of the circle $x^2+y^2=26$ at the point $(5,1)$ :
1) $x+5 y=26$
2) $5 x+y=26$
3) $x+5 y=\sqrt{26}$
4) $5 x+y=\sqrt{26}$

Solution
Here $x_1=5, y_1=1$
So the equation of a tangent to a given circle at $(5,1)$ is

$\mathrm{xx}_1+\mathrm{yy}_1=26$

Thus, we get $5 x+y=26$
Hence, the answer is the option (2).

Example 3: The tangent to the circle $C_1: x^2+y^2-2 x-1=0$ at the point $(2,1)$ cuts off a chord of length 4 from a circle $C_2$ whose centre is ( $3,-2$ ). The radius of $C_2$ is.
1) 2
2) $\sqrt{2}$
3) 3
4) $\sqrt{6}$

Solution
The equation of tangent on $C_1$ at $(2,1)$ is

$\begin{aligned}
& x x_1+y y_1-\left(x+x_1\right)-1=0 \\
& 2 x+y-(x+2)-1=0 \\
& x+y=3
\end{aligned}$

It cuts off the circle $C_2$;
Distance of the line from the centre $(3,-2)$ of $\mathrm{C}_2$

$\Rightarrow\left|\frac{3-2-3}{\sqrt{2}}\right|=\sqrt{2}$

Length of chord $=4$

Using concepts of intercepts

$\begin{aligned}
& r^2=(\text { chord length/2 })^2+(\text { distance of line from centre })^2 \\
& \Rightarrow r^2=4+2=6 \\
& r=\sqrt{6}
\end{aligned}$

Example 4: A line meets the coordinate axes in A and B. A circle is circumscribed about the triangle OAB. If d₁ and d₂ are the distances of the tangent to the circle at the origin O from the points A and B respectively, then the diameter of the circle is:

1) $\frac{2 d_1+d-2}{2}$
2) $\frac{d_1+2 d_2}{2}$
3) $d_1+d_2$
4) $\frac{d_1 d_2}{d_1+d_2}$

Solution
Condition of tangency -
Length of perpendicular from centre of circle $(0,0)$ on the line $y=m x+c$
is Radius of circle

$\begin{aligned}
& \text { i.e., } \frac{|c|}{\sqrt{1+m^2}}=a \\
& \mathrm{c}= \pm \mathbf{a} \sqrt{1+\mathbf{m}^2}
\end{aligned}$

- wherein

If $y=m x+c$ is a tangent to the circle $x^2+y^2=a^2$
Equation of circum circle of triangle $O A B$

$x^2+y^2-a x-b y=0$
Equation of tangent at origin ax + by = 0.

$\begin{aligned} d_1 & =\frac{\left|a^2\right|}{\sqrt{a^2+b^2}} \text { and } d_2=\frac{\left|b^2\right|}{\sqrt{a^2+b^2}} \\ d_1+d_2 & =\sqrt{a^2+b^2}\end{aligned}$

= diameter

Example 5: A circle of radius r such that both coordinates of its centre are positive, touches the x -axis and line $3 y-4 x=0$, then the equation of circle is
1) $x^2+y^2+4 r x-2 r y+4 r^2=0$
2) $x^2+y^2-4 r x-r y+4 r^2=0$
3) $x^2+y^2-2 r x-2 r y+r^2=0$
4) $x^2+y^2-4 r x-2 r y+4 r^2=0$

Solution

Answer (4)

Circle touches x -axis so its y -co-ordinate of $x-$ axis $=$ radius $=r$ Now it touches line $3 y-4 x=0$
So,

$\begin{aligned}
& \frac{3 r-4 h}{5}= \pm r \\
\Rightarrow & 3 r-4 h=5 r \text { or } 3 r-4 h=-5 r \\
\Rightarrow & h=\frac{-2 r}{4} \quad h=2 r
\end{aligned}$

h should be positive
So, $h=2 r$
Now equation circle is

$\begin{aligned}
& (x-2 r)^2+(y-r)^2=r^2 \\
& x^2+y^2-4 r x-2 r y+4 r^2=0
\end{aligned}$

Summary

The equation of the tangent to a circle is a fundamental concept in analytic geometry that provides insights into the geometric properties and spatial relationships involving circles. Whether deriving the tangent from an external point or at a specific point on the circle, understanding these equations is essential for solving geometric problems and applying these concepts in various fields, including optimization, computer graphics, and engineering design. By mastering the methods for finding tangents, one can effectively analyze and utilize the properties of circles in both theoretical and practical applications.