 # Segment BC is Tangent to Circle A at Point B. What is the length of segment BC? 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:

$(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2$

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:

$(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2$

$(AC)^2=(BC)^2 + (AB)^2$

$(15)^2=(4)^2+(AB)^2$

$225=16+(AB)^2$

$Perpendicular = 14.45cm$