Figure 1

In this question, we have to find the** length of line segment** BC which is **tangent at a point** A to the **circle** with the **center at point** B.

The basic concept behind this question is the sound knowledge of **trigonometry**, the **equation of a circle**, the** Pythagoras theorem**, and its application.

**Pythagoras’ theorem** states that the **sum** of the **square of the base** and **perpendicular** of a **right-angled triangle** is equal to the **square of its hypotenuse.**

According to **the Pythagoras theorem**, we have the following formula:

\[ (Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2 \]

## Expert Answer

As we know, a** tangent line** is a line that makes $90^°$. So a line tangent to the circle will be at $90^°$. As point $A$ is the **center of the circle** then line $AB$ will be **perpendicular** to line $BC$, and we can conclude that **angle** $B$ would be a **right angle** which is $90^°$.

Thus, we can write:

\[ AB\bot\ BC\ \]

\[ <B = 90^{°} \]

We also know that $AB $ is the **radius of the circle** and as given it is equal to $21$:

\[ AB = 21 \]

As the point $E $ also lies on the **circle**, so we can conclude that **line** $ AE$ will also be considered as the **radius** and we can write it as:

\[ AE = 21 \]

Given in the figure, we have:

\[ EC = 8 \]

\[ AB = 21 \]

We can write that:

\[ AC = AE + EC \]

\[ AC = 21 + 8 \]

\[ AC = 29 \]

It is obvious that the **triangle** $ABC$ is a **right-angled triangle** and we can apply the **Pythagoras theorem** to it.

According to the **Pythagoras theorem**, we can have the following formula:

\[ (Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2 \]

\[ (AC)^2 = (BC)^2 + (AB)^2 \]

Putting the values of $ AB=21$, $ AC =29$ in the above formula, we get:

\[ (29)^2 = (BC)^2 + (21)^2 \]

\[ 841 = BC^2 + 441 \]

\[ 841 -441 = BC^2 \]

\[ BC^2 = 841 -441 \]

\[ BC^2 = 841 -441 \]

\[ BC^2 = 400 \]

Taking **under root** both sides of the equation, we get:

\[ \sqrt BC^2 = \sqrt 400 \]

\[ BC = 20 \]

## Numerical Results

The** length of line segment** $ BC$ which is **tangent at a point** $ A$ to the **circle** with the **center at point** $B$ is:

\[ Length \space of \space segment \space BC = 20\]

## Example

For a **right-angled triangle**, the** base** is $4cm$ and the **hypotenuse** is $15cm$, calculate the **perpendicular** **of the triangle.**

**Solution**

Let us suppose:

\[ hypotenuse = AC = 15cm \]

\[ base = BC = 4cm \]

\[ perpendicular = AB =? \]

According to the **Pythagoras theorem**, we can have the following formula:

\[ (Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2 \]

\[(AC)^2=(BC)^2 + (AB)^2\]

\[(15)^2=(4)^2+(AB)^2 \]

\[ 225=16+(AB)^2 \]

\[ Perpendicular = 14.45cm \]