Question Video: Finding the Parametric Equations of a Diameter of a Circle | Nagwa Question Video: Finding the Parametric Equations of a Diameter of a Circle | Nagwa

# Question Video: Finding the Parametric Equations of a Diameter of a Circle Mathematics • First Year of Secondary School

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Line segment π΄π΅ is the diameter of circle π, where π΄ = (2, 3) and π΅ = (4, 5). Find the parametric equation of the line that passes through the center of the circle with direction vector β©5, 3βͺ.

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### Video Transcript

Line segment π΄π΅ is the diameter of circle π, where π΄ is the point two, three and π΅ is the point four, five. Find the parametric equation of the line that passes through the center of the circle with direction vector five, three.

In this question, weβre given some information about a circle and a line, and we need to use this information to determine the parametric equation of the line. And to do this, letβs start by recalling what we mean by the parametric equations of a line. Itβs a pair of equations of the form π₯ is equal to π₯ sub zero plus π times π and π¦ is equal to π¦ sub zero plus π multiplied by π. Where the value of π is called a scalar, since it can take any value, the point π₯ sub zero, π¦ sub zero lies on the line and the vector π, π is a nonzero vector which is parallel to the line.

And itβs worth noting we can choose any point on the line to be the point π₯ sub zero, π¦ sub zero. And we can choose any nonzero vector parallel to the line to be the vector π, π. All of these will give us equivalent parametric equations of the line. So to determine the parametric equations of this line, we need to determine the coordinates of a point which lies on the line and a vector which lies parallel to the line. And we can do this by first noticing weβre given a direction vector of the line. Weβre told that this is the vector five, three. And since we can recall the direction vectors are nonzero vectors parallel to the line, we can just set our value of π equal to five and our value of π equal to three.

Therefore, to determine the parametric equations of this line, we just need to determine the coordinates of a point which lies on the line. And we can do this by noting weβre told that the line passes through the center of the circle π. So we want to determine the coordinates of the center of this circle. And we can do this by noting the line segment π΄π΅ is a diameter of the circle, and weβre given the coordinates of its endpoints.

Letβs sketch the information weβre given. We have our circle π, our points π΄ two, three and π΅ four, five, which form a diameter of the circle, so it passes through the center of the circle. In particular, we can notice the center of the circle will be the bisector of any of its diameters. Since any line segment from the center of the circle to its circumference will be a radius of the circle, so the length of these two line segments will be equal. And we can find the coordinates of the center of this circle. Letβs call that πΆ by recalling how we find the coordinates of the midpoint given the endpoints of the line segment.

We just find the average of the π₯-coordinates and π¦-coordinates. πΆ will have coordinates two plus four over two, three plus five over two. πΆ has the point three, four. And remember, weβre told that our line passes through the center of the circle πΆ. So we can choose π₯ sub zero to be equal to three and π¦ sub zero to be equal to four in our parametric equations of the line. Substituting π is equal to five, π is equal to three, π₯ sub zero is equal to three, and π¦ sub zero is equal to four into our parametric equations gives us that π₯ is equal to three plus five π and π¦ is equal to four plus three π.

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