# Question Video: Idenitfying the Parametric Equations of a Diameter of a Circle Mathematics

If the line segment 𝐴𝐵 is the diameter of a circle with center 𝑀, where 𝐴 = (3, 2) and 𝐵 = (1, 2), then which of the following are parametric equations for the line that passes through 𝑀 and point (4, 1)? [A] 𝑥 = 2 − 𝑘, 𝑦 = 2 + 2𝑘 [B] 𝑥 = 2 + 2𝑘, 𝑦 = 2 − 𝑘 [C] 𝑥 = 2 − 2𝑘, 𝑦 = 2 − 𝑘 [D] 𝑥 = 2 + 2𝑘, 𝑦 = 2 + 𝑘 [E] 𝑥 = 2 + 2𝑘, 𝑦 = 2

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### Video Transcript

If the line segment 𝐴𝐵 is the diameter of a circle with center 𝑀, where 𝐴 is the point three, two and 𝐵 is the point one, two, then which of the following are parametric equations for the line that passes through 𝑀 and point four, one? Is it option (A) 𝑥 is equal to two minus 𝑘 and 𝑦 is equal to two plus two 𝑘? Option (B) 𝑥 is equal to two plus two 𝑘 and 𝑦 is equal to two minus 𝑘. Option (C) 𝑥 is equal to two minus two 𝑘 and 𝑦 is equal to two minus 𝑘. Option (D) 𝑥 is equal to two plus two 𝑘 and 𝑦 is equal to two plus 𝑘. Or is it option (E) 𝑥 is equal to two plus two 𝑘 and 𝑦 is equal to two?

In this question, we’re given some information about a line. For example, it passes through the center of a circle 𝑀 and it passes through the point four, one. We need to use this information to determine which of five given options are correct parametric equations for this line. And to do this, let’s start by recalling what we mean by the parametric equations of a line. They’re two equations of the form 𝑥 is equal to 𝑥 sub zero plus 𝑎 times 𝑘 and 𝑦 is equal to 𝑦 sub zero plus 𝑏 times 𝑘. 𝑘 is called the parameter of these equations. It can take any value. In particular, if we substitute 𝑘 is equal to zero into these equations, we see that 𝑥 is equal to 𝑥 sub zero and 𝑦 is equal to 𝑦 sub zero. This means the point with coordinates 𝑥 sub zero and 𝑦 sub zero must lie on the line. In fact, we can choose any point on the line to be this point.

Similarly, we can recall that our line must be parallel to the vector 𝐚, 𝐛 And it’s worth recalling we can choose any nonzero vector parallel to the line to be this vector. We can choose any of these values for 𝐚 and 𝐛. This means there’s many different ways of creating parametric equations for a given line, since we can choose any point on the line for 𝑥 sub zero, 𝑦 sub zero and any nonzero vector parallel to the line for the vector 𝐚, 𝐛.

So let’s start by looking at the five options we’re given. We can notice something interesting about the point 𝑥 sub zero, 𝑦 sub zero for these five options. In all five of the given options, the constant in the 𝑥-equation is two and the constant in the 𝑦-equation is also two. This means all five of these options have chosen the point two, two to lie on the line. 𝑥 sub zero is two, and 𝑦 sub zero is also two. So, when we generate our parametric equations, we don’t need to worry about which point we’re going to choose. We’re going to choose the point two, two since all five of the options choose this point.

The difference in these options comes with choosing the vector parallel to the line. So let’s determine how we can find the vector parallel to this line. We can start by noting we’re given two points on the line: 𝑀 and four, one. So the line must run parallel to the vector between the points 𝑀 and four, one. However, we can’t directly use this to find this vector since we’re not directly given the coordinates of point 𝑀. Instead, we need to note that 𝑀 is the center of a circle where the line segment 𝐴𝐵 is a diameter. And we’re given the coordinates of points 𝐴 and 𝐵. 𝐴 is three, two, and 𝐵 is one, two.

We can use this information to determine the coordinates of the center of the circle 𝑀. Let’s start by sketching our circle and the points 𝐴 and 𝐵 with our diameter, the line segment 𝐴𝐵. Since the line segment 𝐴𝐵 is a diameter, we can note that the line segment 𝑀𝐵 and the line segment 𝑀𝐴 must be radii of the circle, which means they have the same length. In particular, this means that 𝑀 is the midpoint of points 𝐴 and 𝐵.

We can recall we can determine the coordinates of a midpoint of two points by averaging their coordinates. In particular, we can note the average of the 𝑥-coordinates of points 𝐴 and 𝐵 is two and the average of the 𝑦-coordinates is two. 𝑀 is the point two, two, and it’s also the point we’re going to use for our parametric equations, which lies on the line. We could also add onto our diagram the point with coordinates four, one. Remember, our line passes through 𝑀 and this point. And in particular, this means the line must be parallel to the vector between these two points. We’ll call this vector 𝐝. And since we have the coordinates of the initial point and terminal point of vector 𝐝, we can use these to find vector 𝐝. We subtract the position vector of the initial point of vector 𝐝 from the position vector of its terminal point. 𝐝 is the vector four, one minus the vector two, two.

And now we just subtract these two vectors by subtracting their corresponding components. We get that 𝐝 is the vector four minus two, one minus two. And we can evaluate this. 𝐝 is the vector two, negative one. And now we can substitute 𝑎 is equal to two, 𝑏 is equal to negative one along with 𝑥 sub zero is two and 𝑦 sub zero is two into our parametric equations to find one set of parametric equations for this line. Doing this, we get that 𝑥 is equal to two plus two 𝑘 and 𝑦 is equal to two minus 𝑘, which we can see is option (B).

Therefore, we were able to show that 𝑥 is equal to two plus two 𝑘 and 𝑦 is equal to two minus 𝑘 is one set of parametric equations of the line passing through the point 𝑀 and four, one, where 𝑀 is the center of the circle with diameter the line segment 𝐴𝐵, where 𝐴 is the point three, two and 𝐵 is the point one, two.