banner

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:

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

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:

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

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

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

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

\[ Perpendicular = 14.45cm \]

5/5 - (16 votes)