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 | Nagwa 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 | Nagwa

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 • First Year of Secondary School

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A circle has center π and radius 13 cm. A line passes through the points π΅, πΆ, and π·, where πΆ and π· are on the circle, π΅ is 25 cm from the point π, and πΆπ΅ = πΆπ·. Calculate the length of line segment πΆπ· and the perpendicular distance π₯ between the line and the point π. Round your answers to 2 decimal places.

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Video Transcript

A circle has center π and radius 13 centimeters. A line passes through the points π΅, πΆ, and π·, where πΆ and π· are on the circle, π΅ is 25 centimeters from the point π, and πΆπ΅ equals πΆπ·. Calculate the length of line segment πΆπ· and the perpendicular distance π₯ between the line and the point π. Round your answers to two decimal places.

Letβs start by sketching an image. We have a circle with center π. And then we have three points π΅, πΆ, and π·. A line passes through all three of these points. But only πΆ and π· are on the circle. If we have a line that looks like this, we know that πΆ and π· lie on the circle. And we know that the distance from πΆ to π· is equal to the distance from π΅ to πΆ. For now, letβs label the distance from πΆ to π΅ and the distance from πΆ to π· as lowercase π. We know that thereβs a line segment from π΅ to π. This distance is 25 centimeters. We want to find the perpendicular distance between the segment πΆπ· and point π. And weβve been given that that is π₯.

So now we have that π is equal to πΆπ·, which is equal to πΆπ΅, and that π₯ is the perpendicular distance from π to πΆπ·. Weβre trying to find both of these values. In order to do this, letβs remember that this circle has a radius of 13 centimeters. This means the distance from πΆ to π is equal to the distance from π· to π. Both of these values are 13 centimeters. This means we can conclude that triangle πΆππ· is an isosceles triangle.

Weβll label the perpendicular intersection between the line that passes through π and the line πΆπ· be point π΄. When we do this, we can say that the line segment ππ΄ bisects the line segment πΆπ· because the height of an isosceles triangle bisects the base. Weβve labeled the distance πΆπ· π. If we have some point that bisects that, then we can say that segment πΆπ΄ is equal to segment π·π΄. And weβll label that as π over two.

From here, weβll have to consider some right triangle values. We have the larger right triangle π΅π΄π. This triangle has its longest side, its hypotenuse, as 25 centimeters. It has a side of length π₯. And then its final side will be made up of π plus π over two, making its third side equal in length to three π over two. We also have a smaller right triangle, triangle πΆπ΄π. The hypotenuse of this triangle is one of the radiis of this circle πΆπ, and itβs 13 centimeters. It shares the side π΄π, which is π₯ in length. Its final side has a length of π over two.

At this point, we want to come up with some equations that will help us solve for π₯ and π. To do this, we can use the Pythagorean theorem, which says π squared equals π squared plus π squared, where π is the hypotenuse and π and π are the other two sides in a right triangle. Substituting what we know for triangle π΅π΄π, we get 25 squared equals π₯ squared plus three π over two squared. And for triangle πΆπ΄π, we have 13 squared equals π₯ squared plus π over two squared.

If we square all the terms we know, we recognize that each of these equations has an π₯ squared term. We can rearrange this first equation by subtracting nine π squared over four from both sides of the equation, giving us π₯ squared is equal to 625 minus nine π squared over four. We can take the value that weβve found for π₯ squared and plug it in to our second equation, which now says 169 equals 625 minus nine π squared over four plus π squared over four. Combining these like terms, negative nine π squared over four plus π squared over four is negative eight π 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 π squared. Dividing both sides of the equation by negative two, and we have π squared is equal to 228. Taking the square root of both sides gives us π 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 weβre dealing with length, weβre only interested in the positive square root value for π.

Remember, weβve already written an equation for π₯ in terms of π: π₯ squared equals 625 minus nine π squared over four. When solving for π₯, we want to use our nonrounded π-value. This maintains our accuracy as weβre only rounding in the final step. When we plug this in our calculator, we find that π₯ 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 πΆπ· is what weβve labeled on our diagram as π. And that means πΆπ· is equal to 15.10 centimeters to two decimal places. And the perpendicular distance π₯ to two decimal places is 10.58 centimeters.

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