### Video Transcript

A line πΏ intersects a circle with center π. Point π΄ lies on πΏ and is inside the circle. If the radius of the circle is eight centimeters, line segment ππ΄ is perpendicular to πΏ, and we set ππ΄ equal to three π₯ minus five centimeters, in what interval does the value of π₯ belong?

Thereβs an awful lot of information going on here. So letβs begin by sketching out what we know about the line and the circle. We have a circle with center π. The radius of the circle is eight centimeters. And we remember that the radius is a line segment that joins the center of the circle to any point on its circumference. We then have a line that intersects the circle. Weβre told that a point π΄ lies on that line such that ππ΄ is perpendicular to the line itself as shown on the diagram. And weβre told that ππ΄ equals three π₯ minus five centimeters. And weβre looking to find a range of values for π₯. So letβs think about what we know about the diagram we have.

First, we can deduce that line segment ππ΄ must be less than eight centimeters. This is because the radius of the circle is eight centimeters. And so, if we continue line segment ππ΄ to join a point on the circumference of the circle, we have the radius. And this is longer than line segment ππ΄. Since ππ΄ is defined to be three π₯ minus five, then we can form an inequality by replacing ππ΄ with this expression: three π₯ minus five is less than eight.

To find one of the bounds of our interval for π₯ then, letβs solve this inequality. We add five to both sides. And that gives us three π₯ is less than 13. Next, we divide through by three. And that gives us π₯ is less than 13 over three. So, in fact, that gives us the upper bound of the interval in which π₯ belongs. It must be less than 13 over three.

But what do we know about the lower bound? Well, we can deduce that ππ΄ must be greater than zero. If ππ΄ was equal to zero, for instance, it would not be a line; itβd simply be a point in space. And ππ΄ is a length, so it canβt be negative. So we need to solve an inequality by replacing ππ΄ with three π₯ minus five. Thatβs three π₯ minus five is greater than zero. This time, weβre going to add five to both sides. And that gives us three π₯ is greater than five. Next, we divide through by three and we get π₯ is greater than five-thirds.

And so, we have a lower and upper bound for the interval in which π₯ belongs. We could represent this using inequality notation as shown, π₯ is greater than five-thirds and less than thirteen thirds. Or we can use set notation. π₯ lies in the open interval from five-thirds to thirteen thirds.