Video: CBSE Class X • Pack 3 • 2016 • Question 22

CBSE Class X • Pack 3 • 2016 • Question 22

02:47

Video Transcript

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

Let’s start by drawing a diagram. Here, we’ve got a circle with center 𝑂 and an external point 𝐴. And the tangents of the circle going through the external point are 𝐴𝐵 and 𝐴𝐶. And what we need to prove is the length 𝐴𝐵 is equal to the length 𝐴𝐶.

Let’s start by drawing in two radii. The radii we have drawn in are 𝑂𝐵 and 𝑂𝐶. So these are the lines going from the center of the circle to the points where the tangent touch the circle. Since 𝑂𝐵 and 𝑂𝐶 are both radii of the circle, they’ll have the same length. Let’s call this length 𝑟.

Next, let’s add in the line 𝐴𝑂. And we can call this length 𝑥. Now, since 𝐴𝐵 and 𝐴𝐶 are both tangents to the circle and 𝑂𝐵 and 𝑂𝐶 are both radii of the circle, this means that angles 𝐴𝐵𝑂 and 𝐴𝐶𝑂 will both be right angles or 90 degrees. And now, we can see that we formed two right-angled triangles. And these are triangle 𝐴𝐵𝑂 and triangle 𝐴𝐶𝑂. And so we can apply Pythagoras’s theorem to both of them.

Starting with the triangle 𝐴𝐵𝑂, Pythagoras’s theorem tells us that the hypotenuse squared will be equal to the sum of the other two sides squared. And the length of the hypotenuse is 𝑥. So 𝑥 squared is equal to the sum of the other two sides squared. So that’s 𝑟 squared plus 𝐴𝐵 squared.

And now in order to get this in terms of 𝐴𝐵, we subtract 𝑟 squared from both sides. This gives us that 𝑥 squared minus 𝑟 squared is equal to 𝐴𝐵 squared. Taking the square root of both sides gives us that 𝐴𝐵 is equal to the square root of 𝑥 squared minus 𝑟 squared.

Now, we can consider the right-angled triangle 𝐴𝐶𝑂 and apply Pythagoras’s theorem to it. We have that the hypotenuse squared, so that’s 𝑥 squared, is equal to the sum of the other two sides squared. So that’s 𝑟 squared plus 𝐴𝐶 squared.

Now, in order to get this in terms of 𝐴𝐶, we subtract 𝑟 squared from both sides and then take the square root of both sides, giving us that 𝐴𝐶 is equal to the square root of 𝑥 squared minus 𝑟 squared. Now, we can see that 𝐴𝐵 is equal to the square root of 𝑥 squared minus 𝑟 squared. And 𝐴𝐶 is also equal to the square root of 𝑥 squared minus 𝑟 squared.

And since 𝐴𝐵 and 𝐴𝐶 are equal to the same thing since we have shown that 𝐴𝐵 is equal to 𝐴𝐶, we’ve now completed our proof.

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