Equation of a Normal to a Circle

Equation of the Normal 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)

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 $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ to a circle $\mathrm{S}=\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$ 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 $(\mathrm{CP})$ 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 $: x^2+y^2=a^2$ and let the tangent to it be : $y=m x+a \sqrt{\left(1+m^2\right)}$
As the tangent passes through point $P\left(x_1, y_1\right)$ lying out side the circle then, $\mathrm{y}_1=\mathrm{mx}_1+\mathrm{a} \sqrt{\left(1+\mathrm{m}^2\right)}$

$\left(y_1-m x_1\right)^2=a^2\left(1+m^2\right)$

or, $\left(x_1^2-a^2\right) m^2-2 m x_1 y_1+y_1^2-a^2=0$
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

$x^2+y^2+2 g x+2 f y+c=0 \text { is } \sqrt{x_1^2+y_1^2+2 g x_1+2 f y_1+c}$

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

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

Example 1: Let the normals at all the points on a given curve pass through a fixed point ( $\mathrm{a}, \mathrm{b})$. If the curve passes through $(3,-3)$ and $(4,-2 \sqrt{2})$ and given that $a-2 \sqrt{2} b=3$, then $\left(a^2+b^2+a b\right)$ is equal to $\qquad$
1) 9
2) 10
3) 5
4) 7

Solution
All normals of a circle pass through center Radius $=\mathrm{CA}=\mathrm{CB}$

$\begin{aligned}
& \mathrm{CA}^2=\mathrm{CB}^2 \\
& (a-3)^2+(b+3)^2=(a-4)^2+(b-2 \sqrt{2})^2 \\
& a+(3-2 \sqrt{2}) b=3 \\
& a-2 \sqrt{2} b+3 b=3
\end{aligned}$
$\text { given that } a-2 \sqrt{2} b=3$

from above, we get $a=3$ and $b=0$

$a^2+b^2+a b=9$
Hence, the answer is the option (1).

Example 2: Let the lines $(2-i) \bar{z}=(2+i) \bar{z}-4 i=0,\left(\right.$ here $\left.i^2=-1\right)$ be normal to a circle C . if the line $i z+\bar{z}+1+i=0$ is tangent to the circle C , then its radius is:
1) $\frac{1}{2 \sqrt{2}}$
2) $3 \sqrt{2}$
3) $\frac{3}{2 \sqrt{2}}$
4) $\frac{3}{\sqrt{2}}$

Solution

Let $z=x+i y$

$\begin{aligned}
& (i)(2-\mathrm{i}) \mathrm{z}=(2+\mathrm{i}) \overline{\mathrm{z}} \\
& (2-\mathrm{i})(x+i y)=(2+\mathrm{i})(x-i y) \\
& 2 x+2 i y-i x+y=2 x-2 i y+i x+y \\
& y=\frac{x}{2}
\end{aligned}$

$\begin{aligned}
& (i i)(2+i) z+(i-2) \bar{z}-4 i=0 \\
& (2+i)(x+i y)+(i-2)(x-i y)-4 i=0 \\
& (2 x+i x-y+2 i y)+(i x-2 x+y+2 i y)-4 i=0 \\
& x+2 y=2
\end{aligned}$

$\begin{aligned}
& \text { (iii) } \mathrm{iz}+\overline{\mathrm{z}}+1+\mathrm{i}=0 \\
& (i x-y)+(x-i y)+1+i=0
\end{aligned}$

$\mathrm{Eq}^{\mathrm{n}}$ of tangent $\mathrm{x}-\mathrm{y}+1=0$
Solving (i) and (ii)

$\mathrm{x}=1, \mathrm{y}=\frac{1}{2}$
Now, $p=r \Rightarrow\left|\frac{1-\frac{1}{2}+1}{\sqrt{2}}\right|=r$

$\Rightarrow r=\frac{3}{2 \sqrt{2}}$
Hence, the answer is the option (3).

Example 3: The line $2 x-y+1=0$ is a tangent to the circle at the point $(2,5)$ and the centre of the circle lies on $x-2 y=4$. Then the radius of the circle is :
1) $5 \sqrt{3}$
2) $4 \sqrt{5}$
3) $5 \sqrt{4}$
4) $3 \sqrt{5}$

Solution

Slope of tangent is $\mathrm{m}_1=2$
Hence slope of normal from point $A(2,5)$ to the centre of circle is

$m_2=\left(\frac{5-\frac{(h-4)}{2}}{2-h}\right)$
Tangent is perpendicuar to the normal

$\begin{aligned}
& m_1 m_2=-1 \\
& \left(\frac{h-\frac{(h-4)}{2}}{2-h}\right)(2)=-1 \\
& h=8
\end{aligned}$

Center $(8,2)$
Radius $\left.=\sqrt{(8-2)^2+(2-5)^2}=3 \sqrt{5}\right)$

Example 4: Find the equation of the normal to the circle $x^2+y^2+2 x+4 y-5=0$ at $(-2,1)$
1) $3 x+y+5=0$
2) $x+5 y+9=0$
3) $5 x-y+9=0$
4) $4 x+y+7=0$

Solution
The given circle is $x^2+y^2+2 x+4 y-5=0$
Its centre is $(-1,-2)$
So the normal passes through $(-2,1)$ and $(-1,-2)$
The slope of this line is -3
Equation of normal

$\begin{aligned}
& (y+2)=-3(x+1) \\
& 3 x+y+5=0
\end{aligned}$
Hence, the answer is the option (1).

Example 5: Equation of normal at the point with parameter $\alpha=\frac{\pi}{3}$ on the circle $x^2+y^2=4$ is ?
1) $y=\sqrt{3} x$
2) $y=-\sqrt{3} x$
3) $y=x$
4) None of these

Solution

The point is

$(a \cos \theta, a \sin \theta)=\left(2 \cos \left(\frac{\pi}{3}\right), 2 \sin \left(\frac{\pi}{3}\right)\right)=(1, \sqrt{3})$
Centre of the circle is $(0,0)$So the normal is the line passing through $(0,0)$ and $(1, \sqrt{3})$
Equation of normal

$\begin{aligned}
& (y-0)=\frac{\sqrt{3}-0}{1-0}(x-0) \\
& y=\sqrt{3} x
\end{aligned}$
Hence the answer is the option (1)]

Summary

The equation of the normal to a circle is a critical concept in geometry that provides insights into the spatial relationships involving circles and lines. By understanding how to derive and use the normal's equation, one can effectively solve various geometric problems and apply these concepts in fields such as computer graphics, engineering, and optimization. Mastery of these methods enhances one's ability to analyze and utilize geometric properties in both theoretical and practical applications.