# Question Video: Finding the Length of the Secant Using the Application of Similarity in a Circle Mathematics • 11th Grade

Given that π΄π΅ is tangent to the circle and that π· and πΆ lie on the circle, determine the value of π₯ to the nearest tenth.

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### Video Transcript

Given that π΄π΅ is tangent to the circle and that π· and πΆ lie on the circle, determine the value of π₯ to the nearest tenth.

So letβs look at this diagram carefully. Weβre told that the line π΄π΅ is tangent to the circle. The line π΅π· intersects the circle in two places: the points πΆ and π·. And therefore, the line π΅π· is a secant. Weβve been given the length π΄π΅ explicitly β itβs 21 centimeters β and the length π΅π· in terms of π₯ and asked to determine its value.

In order to do so, we need to record the relationship that exists between the length of a tangent and the length of a secant that intersect at a point. The relationship is this: if a tangent and the secant intersect outside a circle, then the square of the measure of the tangent is equal to the product of the measures of the secant and its external secant segment.

Now, thatβs quite a long statement. So letβs take it step by step in order to understand what it means in this question. Firstly, it says if a tangent and a secant intersect outside a circle, well, ours do. The line π΄π΅ and the line π΅π· intersect at the point π΅, which is outside the circle. Next, it talks about the square of the measure of the tangent. So in our question, that is π΄π΅ squared.

Next, weβre told that this is equal to the product of the measures of the secant and its external secant segment. The secant is the full line π΅π·. The external secant segment is just a portion of the secant outside the circle. So itβs the segment π΅πΆ.

So we have the equation π΄π΅ squared is equal to π΅π· multiplied by π΅πΆ. Letβs substitute in some values or expressions for each of these parts. π΄π΅ squared is 21 squared, π΅π· is the full length of that line: so itβs π₯ plus π₯ plus two, and then π΅πΆ is equal to π₯. Letβs simplify this equation. 21 squared is equal to 441. And on the right-hand side, the bracket simplifies to two π₯ plus two.

Next, letβs expand the bracket on the right-hand side. We now have 441 is equal to two π₯ squared plus two π₯. Finally, letβs collect all terms on the right-hand side of the equation. To do so, we need to subtract 441 from both sides. So we now have the equation two π₯ squared plus two π₯ minus 441 is equal to zero.

Now, this is a quadratic equation. And it certainly isnβt one that easily factorizes. So in order to solve it, we need to use the quadratic formula. The quadratic formula tells us that the solutions to the equation ππ₯ squared plus ππ₯ plus π equals zero are given by π₯ is equal to negative π plus or minus the square root of π squared minus four ππ all over two π.

For our equation, π and π are both equal to two and π is equal to negative 441. So substituting our values of π, π, and π into the quadratic formula gives π₯ is equal to negative two plus or minus the square root of two squared minus four multiplied by two multiplied by negative 441 all over two multiplied by two. Now, if you use your calculator in order to evaluate this, it will give two possibilities for the positive and the negative square root: either 14.3576 or negative 15.3576.

Remember that π₯ represents a length. Itβs the length of the segment π΅πΆ. And therefore, it must take a positive value. So the solution that we need is the positive one, 14.3576. Weβre also asked to give our answer to the nearest tenth. So we need to round it. The value of π₯ to the nearest tenth is 14.4.

Remember the key fact that we used in this question: if a tangent and a secant intersect outside a circle, then the square of the measure of the tangent is equal to the product of the measures of the secant and its external secant segment.