EQUATION OF A CIRCLE WHEN EXTREMITIES OF A DIAMETER ARE GIVEN

Equation of a circle when the extremities of the diameter are given.

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

Here (x1, y1) and (x2, y2) are the extremities or end points of the diameter.

Example 1 :

Find the equation of the circle if (2, -3) and (3, 1) are the extremities of a diameter.

Solution :

Here (x1, y1)  ==> (2, -3) and (x2, y2) ==> (3, 1)

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

(x - 2) (x - 3) + (y - (-3)) (y - 1)  =  0

(x - 2) (x - 3) + (y + 3) (y - 1)  =  0

x2 - 3x - 2x + 6 + y2 - y + 3y - 3  =  0

x2 - 5x + y2 + 2y + 6 - 3  =  0

x2 + y2  - 5x + 2y + 3  =  0

Example 2 :

Find the equation of the circle described on the line joining the points (1, 2) and (2, 4) as its diameter.

Solution :

Here (x1, y1)  ==> (1, 2) and (x2, y2) ==> (2, 4)

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

(x - 1) (x - 2) + (y - 2) (y - 4)  =  0

x2 - 2x - x + 2 + y2 - 4y - 2y + 8  =  0

x2 - 3x + 2 + y2 - 6y + 8  =  0

x2 + y2 - 3x - 2y + 2 + 8  =  0

x2 + y2 - 3x - 2y + 10  =  0

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