Video: CBSE Class X • Pack 4 • 2015 • Question 24

CBSE Class X • Pack 4 • 2015 • Question 24

04:54

Video Transcript

Prove that the lengths of the tangents drawn from an external point to a circle are equal.

For a question like this, it’s always helpful to draw a diagram. So we’ll start by drawing a circle, which we’ll call 𝐶, which has a centre of 𝑂. We can also mark the external point to our circle, which is mentioned in the question. And we can label this point 𝑃.

We now need to draw tangents from this external point to the circle. And we can remind ourselves the definition of a tangent. The tangent to a circle is a line which intersects the circle at exactly one point. In other words, the line just touches the perimeter of the circle.

Looking at our diagram, we can draw two lines which fit this description and these are the tangents which we’ll work with. These tangents connect our circle 𝐶 with the point 𝑃. We can also label the points at which these tangents touch the circle as 𝐴 and 𝐵. This allows us to refer to the tangents as 𝐴𝑃 and 𝐵𝑃.

Now, the question is asking us to prove that the lengths of these two tangents are equal. And in order to do so, we’ll need to make a few more geometric constructions. To begin, we can connect the centre of our circle with the two points 𝐴 and 𝐵. These two lines that we have just drawn — 𝑂𝐴 and 𝑂𝐵 — are both radii of the circle since they connect the centre 𝑂 to the circumference via a straight line.

Given this information, we’re able to say that the length 𝑂𝐴 is equal to the length 𝑂𝐵. And this is of course equal to the radius of the circle 𝑟. Now that we’ve drawn these lines, we can use our circle theorems to add more information to the diagram. You may be familiar with the fact that when a tangent and a radius of the same circle are connected, they will always be perpendicular.

Looking at our diagram, we can see that this situation occurs twice at the points 𝐴 and 𝐵. We know that perpendicular lines are at 90 degrees to each other. And we’re, therefore, able to mark on a 90-degree angle between the lines 𝑂𝐴 and 𝐴𝑃 and between the lines 𝑂𝐵 and 𝐵𝑃. Now that we’ve marked angle 𝑂𝐴𝑃 and 𝑂𝐵𝑃 as 90 degrees, we can draw one final line on our diagram which connects the centre of the circle 𝑂 with the external point 𝑃.

In doing this, we have created two connected triangles: 𝑂𝐴𝑃 and 𝑂𝐵𝑃. We can see that both of these triangles are right-angled triangles and they share a common side which is their hypotenuse 𝑂𝑃. We now have that triangles 𝑂𝐴𝑃 and 𝑂𝐵𝑃 are both right-angled triangles. They share the hypotenuse and they have one other side 𝑂𝐴 and 𝑂𝐵, respectively, which is equal in length.

This information is enough for us to conclude that triangles 𝑂𝐴𝑃 and 𝑂𝐵𝑃 are congruent using RHS congruency. RHS congruency says that if the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and one side of another right-angled triangle, then these two triangles are congruent.

Now, we know that congruent triangles have the same corresponding angles and the same corresponding side lengths. This means that the third and final side lengths of our triangles 𝐴𝑃 and 𝐵𝑃, respectively, must also be equal. Since the lines 𝐴𝑃 and 𝐵𝑃 are the tangents, we have therefore answered the question and we have proved that the tangents are equal.

As a quick aside, if you didn’t know the RHS congruency rule, there is also another fairly straightforward way you can solve this problem. We can do this using a familiar tool, which is Pythagoras’s theorem. This tells us that for a right-angled triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.

In our case, for both of our right-angled triangles, the length of the hypotenuse is 𝑂𝑃. For the triangle 𝑂𝐴𝑃, we can, therefore, say that 𝑂𝑃 squared is equal to 𝑂𝐴 squared plus 𝐴𝑃 squared. By looking at triangle 𝑂𝐵𝑃, we can also say that 𝑂𝑃 squared is equal to 𝑂𝐵 squared plus 𝐵𝑃 squared.

Let’s now observe this part of the equation that has been formed. To move forward, we can recall that earlier in our method, we proved that the length 𝑂𝐴 and the length 𝑂𝐵 were the same since they’re both radii of the circle 𝐶. Since this is the case, we can also see that by squaring both terms, 𝑂𝐴 squared is equal to 𝑂𝐵 squared.

Now that we know this, we look back at our equation. And we can see that we can cancel the 𝑂𝐴 squared on the left-hand side of our equation and the 𝑂𝐵 squared on the right-hand side of our equation since both of these are equal. After doing so, the equation that we’re left with is that 𝐴𝑃 squared is equal to 𝐵𝑃 squared. By taking the square root of both sides of this equation, we once again arrived at the conclusion that 𝐴𝑃 is equal to 𝐵𝑃.

We have now answered the question. And we have proved using this alternative method that the two tangents 𝐴𝑃 and 𝐵𝑃 are equal.

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