A line 𝐿 intersects a circle with center 𝑀. Point 𝐴 lies on 𝐿 and is inside the circle. If the radius of the circle is eight centimeters, line segment 𝑀𝐴 is perpendicular to 𝐿, and we set 𝑀𝐴 equal to three 𝑥 minus five centimeters, in what interval does the value of 𝑥 belong?
There’s an awful lot of information going on here. So let’s begin by sketching out what we know about the line and the circle. We have a circle with center 𝑀. 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. We’re told that a point 𝐴 lies on that line such that 𝑀𝐴 is perpendicular to the line itself as shown on the diagram. And we’re told that 𝑀𝐴 equals three 𝑥 minus five centimeters. And we’re looking to find a range of values for 𝑥. So let’s think about what we know about the diagram we have.
First, we can deduce that line segment 𝑀𝐴 must be less than eight centimeters. This is because the radius of the circle is eight centimeters. And so, if we continue line segment 𝑀𝐴 to join a point on the circumference of the circle, we have the radius. And this is longer than line segment 𝑀𝐴. Since 𝑀𝐴 is defined to be three 𝑥 minus five, then we can form an inequality by replacing 𝑀𝐴 with this expression: three 𝑥 minus five is less than eight.
To find one of the bounds of our interval for 𝑥 then, let’s solve this inequality. We add five to both sides. And that gives us three 𝑥 is less than 13. Next, we divide through by three. And that gives us 𝑥 is less than 13 over three. So, in fact, that gives us the upper bound of the interval in which 𝑥 belongs. It must be less than 13 over three.
But what do we know about the lower bound? Well, we can deduce that 𝑀𝐴 must be greater than zero. If 𝑀𝐴 was equal to zero, for instance, it would not be a line; it’d simply be a point in space. And 𝑀𝐴 is a length, so it can’t be negative. So we need to solve an inequality by replacing 𝑀𝐴 with three 𝑥 minus five. That’s three 𝑥 minus five is greater than zero. This time, we’re going to add five to both sides. And that gives us three 𝑥 is greater than five. Next, we divide through by three and we get 𝑥 is greater than five-thirds.
And so, we have a lower and upper bound for the interval in which 𝑥 belongs. We could represent this using inequality notation as shown, 𝑥 is greater than five-thirds and less than thirteen thirds. Or we can use set notation. 𝑥 lies in the open interval from five-thirds to thirteen thirds.