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:
\[ (Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2 \]
\[(AC)^2=(BC)^2 + (AB)^2\]
\[(15)^2=(4)^2+(AB)^2 \]
\[ 225=16+(AB)^2 \]
\[ Perpendicular = 14.45cm \]