Question

Find the equation of the circle if the chord of the circle joining (1,2) and (-3,1) subtends 90° at the centre of the circle. ​

Answer 1

Solution :- The chord joining points A ( 1,2) and B ( -3,1) subtends 90° at the centre of the circle . then AB subtends an angle 45° at point X on the circumference . therefore , the equation of the circle is

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

= +- cot45° [(y-2)(x+3) -(x -1( (y-1)]

or, x² + y² + 2x - 3y - 1 = +-[4y-x-7]

or, x² + y² + 3x & 7y + 6 = 0

and, x² + y² + x + y - 8 = 0

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Answer 2

Answer:

x² + y² + 3x & 7y + 6 = 0 and, x² + y² + x + y - 8 = 0 are the equations of the circle

Step-by-step explanation:

The chord joining points A ( 1,2) and B ( -3,1) subtends 90° at the circle's center. then AB subtends an angle 45° at point X on the circumference. therefore,

the equation of the circle is

(x-x1)(x-x2) + (y-y1)(y-y2) = ±cotα[(y-y1)(x-x2) - (x-x1)(y-y2)]

here α = 90/2 = 45

⇒ (x - 1) ( x + 3) + ( y - 2) ( y - 1) = ±cot45° [(y-2)(x+3) -(x -1( (y-1)]

⇒ (x² + 2x -3 + y² -3y + 2) =± 1[xy + 3y -2x -6 - xy + x + y - 1]

⇒ x² + y² + 2x - 3y - 1 = ±[4y-x-7]

⇒  x² + y² + 2x - 3y - 1 = 4y -x-7  or  x² + y² + 2x - 3y - 1 = - 4y+x+7

⇒ x² + y² + 3x - 7y + 6 = 0  or  x² + y² + x + y - 8 = 0