# Question Video: Finding the Length of a Secant of a Circle and Its Distance from the Centre Given the Lengths of Two Circle Tangents Using Similarity in a Circle Mathematics • 11th Grade

A circle has center 𝑀 and radius 13 cm. A line passes through the points 𝐵, 𝐶, and 𝐷, where 𝐶 and 𝐷 are on the circle, 𝐵 is 25 cm from the point 𝑀, and 𝐶𝐵 = 𝐶𝐷. Calculate the length of line segment 𝐶𝐷 and the perpendicular distance 𝑥 between the line and the point 𝑀. Round your answers to 2 decimal places.

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### Video Transcript

A circle has center 𝑀 and radius 13 centimeters. A line passes through the points 𝐵, 𝐶, and 𝐷, where 𝐶 and 𝐷 are on the circle, 𝐵 is 25 centimeters from the point 𝑀, and 𝐶𝐵 equals 𝐶𝐷. Calculate the length of line segment 𝐶𝐷 and the perpendicular distance 𝑥 between the line and the point 𝑀. Round your answers to two decimal places.

Let’s start by sketching an image. We have a circle with center 𝑀. And then we have three points 𝐵, 𝐶, and 𝐷. A line passes through all three of these points. But only 𝐶 and 𝐷 are on the circle. If we have a line that looks like this, we know that 𝐶 and 𝐷 lie on the circle. And we know that the distance from 𝐶 to 𝐷 is equal to the distance from 𝐵 to 𝐶. For now, let’s label the distance from 𝐶 to 𝐵 and the distance from 𝐶 to 𝐷 as lowercase 𝑎. We know that there’s a line segment from 𝐵 to 𝑀. This distance is 25 centimeters. We want to find the perpendicular distance between the segment 𝐶𝐷 and point 𝑀. And we’ve been given that that is 𝑥.

So now we have that 𝑎 is equal to 𝐶𝐷, which is equal to 𝐶𝐵, and that 𝑥 is the perpendicular distance from 𝑀 to 𝐶𝐷. We’re trying to find both of these values. In order to do this, let’s remember that this circle has a radius of 13 centimeters. This means the distance from 𝐶 to 𝑀 is equal to the distance from 𝐷 to 𝑀. Both of these values are 13 centimeters. This means we can conclude that triangle 𝐶𝑀𝐷 is an isosceles triangle.

We’ll label the perpendicular intersection between the line that passes through 𝑀 and the line 𝐶𝐷 be point 𝐴. When we do this, we can say that the line segment 𝑀𝐴 bisects the line segment 𝐶𝐷 because the height of an isosceles triangle bisects the base. We’ve labeled the distance 𝐶𝐷 𝑎. If we have some point that bisects that, then we can say that segment 𝐶𝐴 is equal to segment 𝐷𝐴. And we’ll label that as 𝑎 over two.

From here, we’ll have to consider some right triangle values. We have the larger right triangle 𝐵𝐴𝑀. This triangle has its longest side, its hypotenuse, as 25 centimeters. It has a side of length 𝑥. And then its final side will be made up of 𝑎 plus 𝑎 over two, making its third side equal in length to three 𝑎 over two. We also have a smaller right triangle, triangle 𝐶𝐴𝑀. The hypotenuse of this triangle is one of the radiis of this circle 𝐶𝑀, and it’s 13 centimeters. It shares the side 𝐴𝑀, which is 𝑥 in length. Its final side has a length of 𝑎 over two.

At this point, we want to come up with some equations that will help us solve for 𝑥 and 𝑎. To do this, we can use the Pythagorean theorem, which says 𝑐 squared equals 𝑎 squared plus 𝑏 squared, where 𝑐 is the hypotenuse and 𝑎 and 𝑏 are the other two sides in a right triangle. Substituting what we know for triangle 𝐵𝐴𝑀, we get 25 squared equals 𝑥 squared plus three 𝑎 over two squared. And for triangle 𝐶𝐴𝑀, we have 13 squared equals 𝑥 squared plus 𝑎 over two squared.

If we square all the terms we know, we recognize that each of these equations has an 𝑥 squared term. We can rearrange this first equation by subtracting nine 𝑎 squared over four from both sides of the equation, giving us 𝑥 squared is equal to 625 minus nine 𝑎 squared over four. We can take the value that we’ve found for 𝑥 squared and plug it in to our second equation, which now says 169 equals 625 minus nine 𝑎 squared over four plus 𝑎 squared over four. Combining these like terms, negative nine 𝑎 squared over four plus 𝑎 squared over four is negative eight 𝑎 squared over four. And eight over four is two.

Next, we subtract 625 from both sides of the equation, which gives us negative 456 is equal to negative two 𝑎 squared. Dividing both sides of the equation by negative two, and we have 𝑎 squared is equal to 228. Taking the square root of both sides gives us 𝑎 equals 15.099 and so on. Rounding that to two decimal places gives us 15.10 centimeters. Even though we took the square root, because we’re dealing with length, we’re only interested in the positive square root value for 𝑎.

Remember, we’ve already written an equation for 𝑥 in terms of 𝑎: 𝑥 squared equals 625 minus nine 𝑎 squared over four. When solving for 𝑥, we want to use our nonrounded 𝑎-value. This maintains our accuracy as we’re only rounding in the final step. When we plug this in our calculator, we find that 𝑥 squared equals 112. And the square root of that value is 10.583 continuing. Rounded to two decimal places, this becomes 10.58 centimeters.

So, to summarize, the length of 𝐶𝐷 is what we’ve labeled on our diagram as 𝑎. And that means 𝐶𝐷 is equal to 15.10 centimeters to two decimal places. And the perpendicular distance 𝑥 to two decimal places is 10.58 centimeters.