# Question Video: Finding the Length of a Tangent to a Circle by Solving Two Linear Equations Mathematics • 11th Grade

Given that π΄πΆ = (2π₯ β 3) cm, find π₯ and π¦ to the nearest thousandth.

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### Video Transcript

Given that π΄πΆ equals two π₯ minus three centimetres, find π₯ and π¦ to the nearest thousandth.

Letβs have a look at this diagram more closely. It consists of two circles with centres π and π and then three lines π΄π΅ and π΄πΆ and π΄π·, all of which start to the common point π΄ outside the circles and then meet a point on their circumference. In fact, the lines π΄π΅, π΄πΆ, and π΄π· are all tangents. π΄π΅ and π΄πΆ are tangents to the circle π. And π΄πΆ and π΄π· are tangents to the circle π.

Weβre asked to determine the values of π₯ and π¦, which are involved in expressions for the lengths of two of these tangents. So what weβre going to be interested in is the relationship that exists between the lengths of these tangents.

We recall then that if two tangents are drawn from the same exterior point to a circle, then they are equal in length. This means that the two tangents drawn from the point π΄ to circle π are equal in length. So we have that π΄π΅ is equal to π΄πΆ. And also the two tangents drawn from the point π΄ to the circle π are equal in length. So we have that π΄πΆ is equal to π΄π·. In fact, all three tangents are equal in length.

Now we can start to form some equations. Weβre given in the diagram that the length of π΄π΅ is 19 centimetres. And in the question, weβre told that the length of π΄πΆ is two π₯ minus three centimetres. So substituting these values or expressions for π΄π΅ and π΄πΆ, we have the equation 19 equals two π₯ minus three. We can solve this equation for π₯ by first adding three to each side, giving 22 is equal to two π₯. Then, we can divide both sides of the equation by two, giving 11 equals π₯. So we found the value of π₯. π₯ is equal to 11.

Now we can form our second equation, which involves π¦. The length of π΄π· is π¦ minus five centimetres. And although the length of π΄πΆ was given as two π₯ minus three centimetres, we know that itβs also equal to π΄π΅, which is 19 centimetres. So we can form the equation 19 is equal to π¦ minus five. To solve for π¦, we just need to add five to each side of this equation, giving 24 is equal to π¦. We found the values of π₯ and π¦. But in the question, it asks us to give these values to the nearest thousandth. Theyβre actually integer values. But if we want to write them to the nearest thousandth, weβll need to include three zeros after the decimal point.

So we have the values of π₯ and π¦. π₯ is equal to 11.000 and π¦ is equal to 24.000. Remember the key result that we used in this question was that if two tangents are drawn from the same exterior point to a circle, then they are equal in length.