Question Video: Finding the Length of a Tangent to a Circle by Solving Two Linear Equations Mathematics

The two circles 𝑀 and 𝑁 are touching externally. The line 𝐹𝐴 is a common tangent to them at 𝐴 and 𝐡 respectively, the line 𝐹𝐢 is a common tangent to them at 𝐢 and 𝐷 respectively. Given that 𝐴𝐡 = 11.01 cm, and 𝐢𝐷 = (𝑦 βˆ’ 5) cm, find π‘₯ and 𝑦.

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Video Transcript

The two circles 𝑀 and 𝑁 are touching externally. The line 𝐹𝐴 is a common tangent to them at 𝐴 and 𝐡, respectively. The line 𝐹𝐢 is a common tangent to them at 𝐢 and 𝐷, respectively. Given that 𝐴𝐡 equals 11.01 centimeters and 𝐢𝐷 equals 𝑦 minus five centimeters, find π‘₯ and 𝑦.

So we have two circles with centers 𝑁 and 𝑀 and two lines 𝐹𝐴 and 𝐹𝐢, which are each tangent to both circles. These two lines intersect at a point 𝐹 outside the circles. And we’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. Let’s consider the smaller circle, the circle with center 𝑁. If the tangents drawn from point 𝐹 to the circle are equal in length, then the length of the line segment 𝐡𝐹 is the same as the line segment 𝐷𝐹.

We can therefore form an equation by setting the expression for the length of 𝐡𝐹 equal to the value we’ve been given for the length of 𝐷𝐹. 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 𝑦, is used in the question to specify the length of the line segment 𝐢𝐷. We’re also told that the length of the line segment 𝐴𝐡 is 11.01 centimeters. Let’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 𝐴 and 𝐢, and they still intersect at the point 𝐹. So we have the equation 𝐴𝐹 is equal to 𝐢𝐹.

We already stated that the length of 𝐡𝐹 was the same as the length of 𝐷𝐹. And so it follows that the length of 𝐴𝐡 is the same as the length 𝐢𝐷. So equating the expression for the length of 𝐢𝐷 with the value for the length of 𝐴𝐡, we have the equation 𝑦 minus five equals 11.01. We can solve this equation for 𝑦 by adding five to each side. And we find that 𝑦 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 𝑁 and once for circle 𝑀, we found the values of π‘₯ and 𝑦. π‘₯ is equal to 14.31, and 𝑦 is equal to 16.01.

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