Equation of the Tangent and Normal to the Circle - Practice Questions & MCQ

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 $P\left(x_1, y_1\right)$ is $x x_1+y y_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0$
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 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{fy}_1
\end{gathered}
$

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

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, $\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}=0$ )

NOTE:
In order to find out the equation of a tangent to any 2nd-degree curve, the following points must be kept in mind:
$x^2$ 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 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 $(\mathrm{a} \cos \theta, \mathrm{a} \sin \theta)$ to a circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ is $\mathbf{x} \cos \theta+\mathrm{y} \sin \theta=\mathbf{a}$
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, $\mathrm{x} \cos \theta+\mathrm{y} \sin \theta=\mathrm{a}$

Slope Form

The equation of the tangent to a circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}_{\text {having slope } \mathrm{m} \text { is } \mathbf{y}=\mathbf{m x} \pm \mathbf{a} \sqrt{\left(\mathbf{1 + \mathbf { m } ^ { 2 } )}\right.} \text {, and point of tangency is }}^{\left( \pm \frac{a m}{\sqrt{\left(1+m^2\right)}}, \mp \frac{a}{\sqrt{\left(1+m^2\right)}}\right) .}$
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 a 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 $(y-k)=m(x-h) \pm a \sqrt{\left(1+m^2\right)}$.

Equation of the Normal to a Circle

A line passing through a point P on the curve which is perpendicular to the tangent at P is called the normal to the curve at P.

For a circle, the normal always passes through the centre of the circle.

Point Form:

The equation of the Normal at the point $P\left(x_1, y_1\right)$ to a circle $S=x^2+y^2+2 g x+2 f y+c=0$ is

$
\frac{x-x_1}{g+x_1}=\frac{y-y_1}{f+y_1}
$

Proof:

As we know that the normal always passes through the centre $C(-g,-f)$ of a circle.
Thus, the equation of the normal at point $P$ to the circle

$
\begin{aligned}
& \mathrm{y}-\mathrm{y}_1 \\
= & \frac{\mathrm{y}_1+\mathrm{f}}{\mathrm{x}_1+\mathrm{g}}\left(\mathrm{x}-\mathrm{x}_1\right) \\
\Rightarrow \quad & \frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{x}_1+\mathrm{g}}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{y}_1+\mathrm{f}}
\end{aligned}
$
This is the equation of the normal (CP) at point P of the circle

Tangent from a Point to the Circle

If a point lies outside of a circle (here point is P ), then two tangents can be drawn from P to the circle. Here, PQ and PR are two tangents.

If a point lies on the circle, then one tangent can be drawn from the point to the circle. If C is the point, then ACB is the tangent

If a point lies inside the circle, then no tangent can be drawn from the point to the circle.

To get equation of the tangents from an external point

Circle is : $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ and let the tangent to it be : $\mathrm{y}=\mathrm{mx}+\mathrm{a} \sqrt{\left(1+\mathrm{m}^2\right)}$ As the tangent passes through point $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ lying out side the circle then, $\mathrm{y}_1=\mathrm{mx}_1+\mathrm{a} \sqrt{\left(1+\mathrm{m}^2\right)}$

$
\left(\mathrm{y}_1-\mathrm{mx}_1\right)^2=\mathrm{a}^2\left(1+\mathrm{m}^2\right)
$

or, $\left(x_1^2-a^2\right) m^2-2 \mathrm{mx}_1 \mathrm{y}_1+\mathrm{y}_1^2-\mathrm{a}^2=0$
Which is quadratic equation in m which gives two value of m .

The tangents are real, imaginary or coincidence that is depends on the value of the discriminant.

If we have real values of m, then we can find the equations of 2 tangents using these slopes and the point P. 

Length of tangent (PT)  from a point to a circle

The length of the tangent from a point $\mathrm{P}\left(x_1, y_1\right)$ to the circle $x^2+y^2+2 g x+2 f y+c=0$ is $\sqrt{x_1^2+y_1^2+2 g x_1+2 f y_1+c}$

In $\triangle \mathrm{PTC}, \mathrm{PT}^2=\mathrm{PC}^2-\mathrm{CT}^2$
Here coordinates of C are $(-\mathrm{g},-\mathrm{f})$
Hence, $\quad \mathrm{PT}^2=\left(\sqrt{\left(\mathrm{x}_1+\mathrm{g}\right)^2+\left(\mathrm{y}_1+\mathrm{f}\right)^2}\right)^2-\left(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\right)^2$
$\Rightarrow \quad \mathrm{PT}=\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}} \quad[\because \mathrm{CT}=$ radius $]$
This expression can also be written as $P T=\sqrt{S_1}$