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 (x1, y1) and (x2, y2) as diameter is (x - x1)(x - x2)  + (y - y1)(y - y2) = 0


First Method:

Let P (x1, y1) and Q (x2, y2) 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 (x1+x22, y1+y22).

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

The radius of the required circle

= 12PQ

= 12√(x1βˆ’x2)2+(y1βˆ’y2)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 = a2.

Therefore, the equation of the required circle is

(x - x1+x22)2 + (y - y1+y22)2 = [12√(x1βˆ’x2)2+(y1βˆ’y2)2 ]2

β‡’ (2x - x1 - x2)2 + (2y - y1 - y2)2 = (x1 - x2)2 + (y1 - y2)2

β‡’ (2x - x1 - x2)2 - (x1 - x2)2 + ( 2y - y1 - y2 )2 - (y1 - y2)2 = 0

β‡’ (2x - x1 - x2 + x1 - x2)(2x - x1 - x2 - x1 + x2) + (2y - y1 - y2 + y1 - y2)(2y - y1 - y2 + y2) = 0

β‡’ (2x - 2x2)(2x - 2x1) + (2y - 2y2)(2y - 2y1) = 0

β‡’ (x - x2)(x - x1) + (y - y2)(y - y1) = 0

β‡’ (x - x1)(x - x2) + (y - y1)(y - y2) = 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 (x1, y1) and Q (x2, y2). 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.

m1 = the slope of the straight line PM = yβˆ’y1xβˆ’x1

m2 = the slope of the straight line PQ = yβˆ’y2xβˆ’x2.

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, yβˆ’y1xβˆ’x1 Γ— yβˆ’y2xβˆ’x2 = -1

β‡’ (y - y1)(y - y2) = - (x - x1)(x - x2

β‡’ (x - x1)(x - x2) + (y - y1)(y - y2) = 0.

This is the required equation of the circle having (x1, y1) and (x2, y2) 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.

● The Circle




11 and 12 Grade Math 

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