Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).

Asked by Pragya Singh | 1 year ago |  144

Solution :-

Let A = (6, -6), B = (3, -7), C = (3, 3) are the points on a circle.

If O is the centre, then OA = OB = OC (radii are equal)

If O = (x, y) then

OA = $$\sqrt{(x – 6)^2 + (y + 6)}^2$$

OB = $$\sqrt{(x – 3)^2 + (y + 7})^2$$

OC = $$\sqrt{(x-3)^2+(y – 3)^2}$$

Choose: OA = OB, we have

After simplifying above, we get -6x = 2y – 14 ….(1)

Similarly: OB = OC

(x – 3)+ (y + 7)2 = (x – 3)2 + (y – 3)2

(y + 7)2 = (y – 3)2

y+ 14y + 49 = y– 6y + 9

20y =-40

or y = -2

Substituting the value of y in equation (1), we get;

-6x = 2y – 14

-6x = -4 – 14 = -18

x = 3

Hence, the centre of the circle located at point (3,-2).

Answered by Abhisek | 1 year ago

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