FIND EQUATION OF CIRCLES WHEN ENDPOINTS OF DIAMETER IS GIVEN

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