# Tangent & Normal to a Circle

### Tangent and Normal to a Circle in Coordinate Geometry:

#### 1. **Tangent to a Circle:**

- A tangent to a circle is a line that touches the circle at exactly one point, termed the point of tangency.
- It is perpendicular to the radius at the point of contact.

#### 2. **Equation of Tangent to a Circle:**

- For a circle with center $(h,k)$ and radius $r$, the equation of the tangent at a point $({x}_{1},{y}_{1})$ on the circle is: $(x-{x}_{1})({x}_{1}-h)+(y-{y}_{1})({y}_{1}-k)={r}^{2}$ where $({x}_{1},{y}_{1})$ are the coordinates of the point of tangency.

#### 3. **Slope of Tangent:**

- The slope of the tangent to a circle at the point of tangency is equal to the negative reciprocal of the slope of the radius at that point.

#### 4. **Normal to a Circle:**

- The normal to a curve at a given point is a line perpendicular to the tangent at that point.
- For a circle, the normal at a point on the circle is the line passing through that point and the circle's center.

#### 5. **Equation of Normal to a Circle:**

- The equation of the normal to a circle at a point $({x}_{1},{y}_{1})$ on the circle with center $(h,k)$ is given by: $(y-{y}_{1})=-\frac{(x-{x}_{1})}{({x}_{1}-h)}$

#### 6. **Properties:**

**Perpendicularity:**The tangent and the radius at the point of tangency are perpendicular.**Normal as Perpendicular Line:**The normal to a circle at a point is perpendicular to the tangent at that point.**Unique Point of Contact:**A tangent and normal intersect the circle at only one point.

#### 7. **Derivation:**

- The equations of the tangent and normal are derived based on the slopes of the radius and perpendicular lines, respectively.

### Example:

Consider a circle with center $(3,4)$ and radius $5$.

#### 1. Equation of Tangent:

- Find the equation of the tangent to the circle at the point $(7,4)$ on the circle.

**Solution:**

Given: Center of the circle $=(3,4)$

Radius $=5$

Point on the circle $=(7,4)$

The equation of the tangent to the circle at a point $({x}_{1},{y}_{1})$ on the circle with center $(h,k)$ and radius $r$ is: $(x-{x}_{1})({x}_{1}-h)+(y-{y}_{1})({y}_{1}-k)={r}^{2}$

Substituting the values: $(x-7)(7-3)+(y-4)(4-4)={5}^{2}$

$(x-7)(4)=25$

$4x-28=25$

$4x=53$

$x=\frac{53}{4}$

So, the equation of the tangent to the circle at $(7,4)$ is $x=\frac{53}{4}$.

#### 2. Equation of Normal:

- Find the equation of the normal to the circle at the point $(7,4)$ on the circle.

**Solution:**

The equation of the normal to a circle at a point $({x}_{1},{y}_{1})$ on the circle with center $(h,k)$ is given by: $(y-{y}_{1})=-\frac{(x-{x}_{1})}{({x}_{1}-h)}$

Substituting the values: $(y-4)=-\frac{(x-7)}{(7-3)}$

$(y-4)=-\frac{(x-7)}{4}$

$4(y-4)=-(x-7)$

$4y-16=-x+7$

$x+4y=23$

So, the equation of the normal to the circle at $(7,4)$ is $x+4y=23$.