In this video, we are going to look at line segment lengths of tangents and secants which intersect at an external point.

Here we have two secants drawn through the circle. We can see that each secant has two line segments. We can remember this using a trick: $ew=ew$, or in other words, $exterior(whole)=exterior(whole)$. So the exterior parts are the segments outside of the circle. The whole represents the entire line from the exterior point to the endpoint inside the circle. Let’s label the exteriors as $a$ and $c$, and the interiors as $b$ and $d$. This would mean that $a(a+b)=c(c+d)$.

For example:
If we have the exteriors as 2 and 3, and the interiors as $x$ and 6, then we need to solve for $x$. Here we have $2(2+x)=3(9)$ $4+2x=27$ $2x=23$ $x=\frac{23}{2}=11.5$
If we needed the whole length of the secant, then it would be 13.5.

If we have a tangent line and a secant, instead of two secants, then we still have to use the same formula. Let’s label the tangent $a$ and leave the secant as $c$ and $d$. This would mean that $a(a)=c(c+d)$ or $a^2=c(c+d)$.

For example:
If the tangent is $x$, and the secant has an exterior of 4 and an interior of 5, then we would have to solve for $x$. Here we have $x(x)=4(9)$ $x^2=36$ $\sqrt{x^2}=\sqrt{36}$ $x=\pm6$

However, $x$ represents a length, and we can’t have a negative measurement for distance. Therefore, we only accept the positive answer and reject the negative. So, our final answer is $x=6$.

## Video-Lesson Transcript

In this lesson, we’ll look at tangent line and secant line relationships.

In this circle, we have two secants drawn. They both intersecting at one external point.

Let’s name some of these line segments.

As you can see the horizontal line, when passed the circle, made two line segments. One inside the circle and the second outside the circle.

Same thing with the slanting line. There is an internal segment and exterior segment.

The relationship between these two line segments is $ew = ew$

In other words, $exterior \times whole = exterior \times whole$

The exterior parts are the segments outside of the circle. The whole represents the entire line. This starts from the exterior point until the endpoint intercepting the circle. If the line continues outside, we will not count that. The whole is just until where it ends inside the circle.

Let’s give some measurements.

The horizontal exterior line segment is $a$, the interior line segment is $b$, the slanting line’s exterior segment is $c$, and its interior segment is $d$. $a \times (a + b) = c \times (c + d)$

Let’s have an example. $a = 2$ $b = x$ $c = 3$ $d = 6$

Following the formula: $ew = ew$ $2 \times (2 + x) = 3 \times (3 + 6)$ $4 + 2x = 3 \times (9)$ $4 + 2x = 27$ $4 - 4 + 2x = 27 - 4$ $2x = 23$ $\dfrac{2x}{2} = \dfrac{23}{2}$ $x = 11.5$

The whole length of the horizal line is $13.5$.

Now, let’s take a look at a different example.

We have a different diagram here.

This time, we have a tangent line and a secant. But we’ll still use the same formula. $ew = ew$

Here, we see that $a$ is the exterior. The whole starts from point $p$ but it ends just when it touches the circle. So, the whole length is $a$.

The slanting line’s exterior is $c$ and the whole length is $c + d$.

So, we have $ew = ew$ $a \times (a) = c \times (c + d)$ $a^2 = c (c + d)$

Let’s look at an example for this one. $a = x$ $c = 4$ $d = 5$

Let’s solve: $ew = ew$ $x (x) = 4 (9)$ $x^2 = 36$ $\sqrt{x^2} = \sqrt{36}$ $x = \pm 6$

But since $x$ represents length, it can’t have a negative value. There is no negative measurement for distance.

We’ll keep $x = 6$.