Equation of a Circle when the Line Segment Joining Two Given Points is a Diameter

We will learn how to find the equation of the circle for which the line segment joining two given points is a diameter.

the equation of the circle drawn on the straight line joining two given points (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)) as diameter is (x - x\(_{1}\))(x - x\(_{2}\))  + (y - y\(_{1}\))(y - y\(_{2}\)) = 0

First Method:

Let P (x\(_{1}\), y\(_{1}\)) and Q (x\(_{2}\), y\(_{2}\)) are the two given given points on the circle. We have to find the equation of the circle for which the line segment PQ is a diameter.

Therefore, the mid-point of the line segment PQ is (\(\frac{x_{1} + x_{2}}{2}\), \(\frac{y_{1} + y_{2}}{2}\)).

Now see that the mid-point of the line segment PQ is the centre of the required circle.

The radius of the required circle

= \(\frac{1}{2}\)PQ

= \(\frac{1}{2}\)\(\mathrm{\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}}\)

We know that the equation of a circle with centre at (h, k) and radius equal to a, is (x - h)\(^{2}\) + (y - k)\(^{2}\) = a\(^{2}\).

Therefore, the equation of the required circle is

(x - \(\frac{x_{1} + x_{2}}{2}\))\(^{2}\) + (y - \(\frac{y_{1} + y_{2}}{2}\))\(^{2}\) = [\(\frac{1}{2}\)\(\mathrm{\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}}\) ]\(^{2}\)

⇒ (2x - x\(_{1}\) - x\(_{2}\))\(^{2}\) + (2y - y\(_{1}\) - y\(_{2}\))\(^{2}\) = (x\(_{1}\) - x\(_{2}\))\(^{2}\) + (y\(_{1}\) - y\(_{2}\))\(^{2}\)

⇒ (2x - x\(_{1}\) - x\(_{2}\))\(^{2}\) - (x\(_{1}\) - x\(_{2}\))\(^{2}\) + ( 2y - y\(_{1}\) - y\(_{2}\) )\(^{2}\) - (y\(_{1}\) - y\(_{2}\))\(^{2}\) = 0

⇒ (2x - x\(_{1}\) - x\(_{2}\) + x\(_{1}\) - x\(_{2}\))(2x - x\(_{1}\) - x\(_{2}\) - x\(_{1}\) + x\(_{2}\)) + (2y - y\(_{1}\) - y\(_{2}\) + y\(_{1}\) - y\(_{2}\))(2y - y\(_{1}\) - y\(_{2}\) + y\(_{2}\)) = 0

⇒ (2x - 2x\(_{2}\))(2x - 2x\(_{1}\)) + (2y - 2y\(_{2}\))(2y - 2y\(_{1}\)) = 0

⇒ (x - x\(_{2}\))(x - x\(_{1}\)) + (y - y\(_{2}\))(y - y\(_{1}\)) = 0

⇒ (x - x\(_{1}\))(x - x\(_{2}\)) + (y - y\(_{1}\))(y - y\(_{2}\)) = 0.

 

Second Method:

equation of a circle when the co-ordinates of end points of a diameter are given

Let the two given points be P (x\(_{1}\), y\(_{1}\)) and Q (x\(_{2}\), y\(_{2}\)). We have to find the equation of the circle for which the line segment PQ is a diameter.

Let M (x, y) be any point on the required circle. Join PM and MQ.

m\(_{1}\) = the slope of the straight line PM = \(\frac{y - y_{1}}{x - x_{1}}\)

m\(_{2}\) = the slope of the straight line PQ = \(\frac{y - y_{2}}{x - x_{2}}\).

Now, since the angle subtended at the point M in the semi-circle PMQ is a right angle.

Now, PQ is a diameter of the required circle.

Therefore, ∠PMQ = 1 rt. angle i.e., PM is perpendicular to QM

Therefore, \(\frac{y - y_{1}}{x - x_{1}}\) × \(\frac{y - y_{2}}{x - x_{2}}\) = -1

(y - y\(_{1}\))(y - y\(_{2}\)) = - (x - x\(_{1}\))(x - x\(_{2}\)

(x - x\(_{1}\))(x - x\(_{2}\)) + (y - y\(_{1}\))(y - y\(_{2}\)) = 0.

This is the required equation of the circle having (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)) as the coordinates of the end points of a diameter.


Note: If the coordinates of the end points of a diameter of a circle given, we can also find the equation of the circle by finding the coordinates of the centre and radius. The centre is the mid-point of the diameter and radius is half of the length of the diameter.




11 and 12 Grade Math 

From Equation of a Circle when the Line Segment Joining Two Given Points is a Diameter to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.