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

The two circles π and π are touching externally. The line πΉπ΄ is a common tangent to them at π΄ and π΅ respectively, the line πΉπΆ is a common tangent to them at πΆ and π· respectively. Given that π΄π΅ = 11.01 cm, and πΆπ· = (π¦ β 5) cm, find π₯ and π¦.

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### Video Transcript

The two circles π and π are touching externally. The line πΉπ΄ is a common tangent to them at π΄ and π΅, respectively. The line πΉπΆ is a common tangent to them at πΆ and π·, respectively. Given that π΄π΅ equals 11.01 centimeters and πΆπ· equals π¦ minus five centimeters, find π₯ and π¦.

So we have two circles with centers π and π and two lines πΉπ΄ and πΉπΆ, which are each tangent to both circles. These two lines intersect at a point πΉ outside the circles. And weβre given values or, in some cases, expressions for the lengths of different segments of these tangents. We recall one of the key circle theorems, which is that tangents drawn from the same external point to a circle are equal in length. Letβs consider the smaller circle, the circle with center π. If the tangents drawn from point πΉ to the circle are equal in length, then the length of the line segment π΅πΉ is the same as the line segment π·πΉ.

We can therefore form an equation by setting the expression for the length of π΅πΉ equal to the value weβve been given for the length of π·πΉ. And we have π₯ minus two is equal to 12.31. We can solve this equation for π₯ by adding two to each side, and we find that π₯ is equal to 14.31. The other value we need to find, which is π¦, is used in the question to specify the length of the line segment πΆπ·. Weβre also told that the length of the line segment π΄π΅ is 11.01 centimeters. Letβs consider then this same circle theorem in relation to the larger circle. In this case, the points at which the tangents meet the circle are π΄ and πΆ, and they still intersect at the point πΉ. So we have the equation π΄πΉ is equal to πΆπΉ.

We already stated that the length of π΅πΉ was the same as the length of π·πΉ. And so it follows that the length of π΄π΅ is the same as the length πΆπ·. So equating the expression for the length of πΆπ· with the value for the length of π΄π΅, we have the equation π¦ minus five equals 11.01. We can solve this equation for π¦ by adding five to each side. And we find that π¦ is equal to 16.01.

So by recalling the key result that tangents drawn from the same external point to a circle are equal in length and then applying this result twice, once for circle π and once for circle π, we found the values of π₯ and π¦. π₯ is equal to 14.31, and π¦ is equal to 16.01.