To find equation of circle, we use the formula
(x - h)^{2} + (y - k)^{2} = r^{2}
Here (h, k) is center of the circle and "r" is radius of circle.
If (x_{1}, y_{1}) and (x_{2}, y_{2}) be end points of diameter of the circle, then
Midpoint of the diameter = center
= (x_{1} + x_{2})/2, (y_{1} + y_{2})/2
Finding distance between center and one endpoint of the diameter, we will get radius
Problem 1 :
A is the point (−3, 1) and B(5, 3).
Find the equation of the circle which has AB as a diameter.
Solution :
AB is the diameter of the circle, Let C be the midpoint and its center of the circle.
C = (-3 + 5)/2, (1 + 3)/2
C = 2/2, 4/2
C = (1, 2)
Distance between C and A :
= √(1 + 3)^{2} + (2 - 1)^{2}
= √4^{2} + 1^{2}
= √16+1
= √17
(x - 1)^{2} + (y - 2)^{2} = √17^{2}
(x - 1)^{2} + (y - 2)^{2} = 17
Using algebraic identities finding expansion, we get
x^{2} - 2x + 1 + y^{2} - 4y + 4 = 17
x^{2} + y^{2} - 2x - 4y + 5 - 17 = 0
x^{2} + y^{2} - 2x - 4y - 12 = 0
Problem 2 :
A is the point (-4, 2) and B is (6, -4). Find the equation of the circle which has AB as a diameter.
Solution :
AB is the diameter of the circle, Let C be the midpoint and its center of the circle.
C = (-4 + 6)/2, (2 + (-4))/2
C = 2/2, -2/2
C = (1, -1)
Distance between C and A :
= √(1 + 4)^{2} + (-1 - 2)^{2}
= √5^{2} + (-3)^{2}
= √25+9
= √34
(x - 1)^{2} + (y + 1)^{2} = √34^{2}
(x - 1)^{2} + (y + 1)^{2} = 34
Using algebraic identities finding expansion, we get
x^{2} - 2x + 1 + y^{2} + 2y + 1 = 34
x^{2} + y^{2} - 2x + 2y + 2 - 34 = 0
x^{2} + y^{2} - 2x + 2y - 32 = 0
Problem 3 :
P is the point (-5, 3) and Q is (5, -21). Find the equation of the circle which has PQ as diameter.
Solution :
Mid point of PQ = (-5 + 5)/2, (3 - 21)/2
= (0, -18/2)
= C(0, -9)
Distance between C and P :
= √(0 + 5)^{2} + (-9 - 3)^{2}
= √5^{2} + (-12)^{2}
= √25 + 144
= √169
= 13
(x - 0)^{2} + (y + 9)^{2} = 13^{2}
Using algebraic identities finding expansion, we get
x^{2} + y^{2} + 18y + 81 = 169
x^{2} + y^{2} + 18y + 81- 169 = 0
x^{2} + y^{2} + 18y - 88 = 0
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