FIND EQUATION OF CIRCLES WHEN ENDPOINTS OF DIAMETER IS GIVEN

To find equation of circle, we use the formula

(x - h)2 + (y - k)2 = r2

Here (h, k) is center of the circle and "r" is radius of circle.

If (x1, y1) and (x2, y2) be end points of diameter of the circle, then

Midpoint of the diameter = center

= (x1 + x2)/2, (y1 + y2)/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

√42 + 12

√16+1

√17

(x - 1)2 + (y - 2)2√172

(x - 1)2 + (y - 2)2 = 17

Using algebraic identities finding expansion, we get

x2 - 2x + 1 + y2 - 4y + 4 = 17

x2 + y2 - 2x - 4y + 5 - 17 = 0

x2 + y2 - 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

√52 + (-3)2

√25+9

√34

(x - 1)2 + (y + 1)2 = √342

(x - 1)2 + (y + 1)2 = 34

Using algebraic identities finding expansion, we get

x2 - 2x + 1 + y2 + 2y + 1 = 34

x2 + y2 - 2x + 2y + 2 - 34 = 0

x2 + y2 - 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

√52 + (-12)2

√25 + 144

√169

= 13

(x - 0)2 + (y + 9)2 = 132

Using algebraic identities finding expansion, we get

x2 + y2 + 18y + 81 = 169

x2 + y2 + 18y + 81- 169 = 0

x2 + y2 + 18y - 88 = 0


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