# EQUATION OF A CIRCLE WITH END POINTS OF DIAMETER

To find equation of the circle with endpoints of diameter, we use the formula given below. Equation of circle :

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

Problem 1 :

Ends of a diameter : (-9, 5) and (1, 1)

Solution :

Now we have to consider the given points as (x1, y1) and (x2, y2). So the values of x1 = -9, y1 = 5, x2 = 1 and y2 = 1

Equation of circle :

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

(x + 9) (x - 1) + (y - 5) (y - 1) = 0

x2 - x + 9x - 9 + y2 - y - 5y + 5 = 0

x2 + 8x - 4 + y2 - 6y = 0

So, the required equation of a circle

x2 + y2 + 8x - 6y - 4 = 0

Problem 2 :

Ends of a diameter : (6, 8) and (16, 6)

Solution :

Here x1 = 6, y1 = 8, x2 = 16 and y2 = 6

Equation of circle :

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

(x - 6) (x - 16) + (y - 8) (y - 6) = 0

x2 - 16x - 6x + 96 + y2 - 6y - 8y + 48 = 0

x2 - 22x + 144 + y2 - 14y = 0

So, the required equation of a circle

x2 + y2 - 22x - 14y + 144 = 0

Problem 3 :

Ends of a diameter : (19, 5) and (-5, 1)

Solution :

Here x1 = 19, y1 = 5, x2 = -5 and y2 = 1

(x - 19) (x + 5) + (y - 5) (y - 1) = 0

x2 + 5x - 19x - 95 + y2  - y - 5y + 5 = 0

x2 - 14x - 90 + y2 - 6y = 0

So, the required equation of a circle

x2 + y2 - 14x - 6y - 90 = 0

Problem 4 :

Ends of a diameter : (11, -5) and (11, -7)

Solution :

Here x1 = 11, y1 = -5, x2 = 11 and y2 = -7

(x - 11) (x - 11) + (y + 5) (y + 7) = 0

x2  - 11x  - 11x + 121 + y2 + 7y + 5y  + 35 = 0

x2 - 22x + 156 + y2  + 12y = 0

So, the required equation of a circle

x2 - 22x + 156 + y2  + 12y = 0.

Problem 5 :

Given a circle with (5, 1) and (3, -1) as the endpoints of the diameter.

Solution :

Here x1 = 5, y1 = 1, x2 = 3 and y2 = -1

(x - 5) (x - 3) + (y - 1) (y + 1) = 0

x2 - 3x - 5x + 15 + y2 + y - y - 1 = 0

x2 - 8x + 14 + y2 = 0

So, the required equation of a circle x2 + y2 - 8x + 14 = 0.

Problem 6 :

Given a circle with (2, 1) and (6, -3) as the endpoints of the diameter.

Solution :

Here x1 = 2, y1 = 1, x2 = 6 and y2 = -3

(x - 2) (x - 6) + (y - 1) (y + 3) = 0

x2 - 6x - 2x + 12 + y2 + 3y - y - 3 = 0

x2 - 8x + 9 + y2 + 2y = 0

So, the required equation of a circle

x2 + y2 - 8x + 2y + 9 = 0.

Problem 7 :

Given a circle with (4, -3) and (2, 1) as the endpoints of the diameter.

Solution :

Here x1 = 4, y1 = -3, x2 = 2 and y2 = 1

(x - 4) (x - 2) + (y + 3) (y - 1) = 0

x2 - 2x - 4x + 8 + y2 - y + 3y - 3 = 0

x2 - 6x + 5 + y2 + 2y = 0

So, the required equation of a circle

x2 + y2 - 6x + 2y + 5 = 0.

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