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the center of the circle is (3,–2)
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hey there,
let O( x , y) is the centre of a circle from which given points A(6 , -6) , B( 3, -7) and C ( 3, 3) .
use here concept of circle ,
radius of circle always constant for all conditions .
here ,
OA = OB = OC =radius
hence, OA² = OB² = OC²
use distance formula,
{(x -6)² +(y +6)² } = OA²
{(x -3)² + (y +7)²} = OB²
{(x -3)² + (y -3)² = OC²
now,
OA² = OB²
{(x -6)² + (y+6)²} = {(x -3)² + (y +7)²}
(x-6)² -(x -3)² = (y+7)²- (y +6)²
(x -6-x +3)(x -6+x -3)=(y+7-y-6)(y +7+y+6)
(-3)(2x -9) =(1)(2y+13)
-6x +27 = 2y +13
2y +6x = 14 ----------(1)
OA² = OC²
(x -6)² + (y +6)²= (x -3)² + ( y-3)²
(x -6)² -(x -3)² = (y-3)² -(y +6)²
(x-6-x +3)(x -6+x -3) =(y-3-y-6)(y-3+y+6)
(-3)(2x -9) = (-9)(2y +3)
2x -9 = 6y +9
2x -6y = 18
x - 3y = 9 ------(2)
from equaion (1) and (2)
2y +6(3y +9) = 14
2y + 18y +54 = 14
20y = -40
y = -2
put this in equation (2)
x = 3
hence, centre of circle =( 3, -2)
Hope this helps!
let O( x , y) is the centre of a circle from which given points A(6 , -6) , B( 3, -7) and C ( 3, 3) .
use here concept of circle ,
radius of circle always constant for all conditions .
here ,
OA = OB = OC =radius
hence, OA² = OB² = OC²
use distance formula,
{(x -6)² +(y +6)² } = OA²
{(x -3)² + (y +7)²} = OB²
{(x -3)² + (y -3)² = OC²
now,
OA² = OB²
{(x -6)² + (y+6)²} = {(x -3)² + (y +7)²}
(x-6)² -(x -3)² = (y+7)²- (y +6)²
(x -6-x +3)(x -6+x -3)=(y+7-y-6)(y +7+y+6)
(-3)(2x -9) =(1)(2y+13)
-6x +27 = 2y +13
2y +6x = 14 ----------(1)
OA² = OC²
(x -6)² + (y +6)²= (x -3)² + ( y-3)²
(x -6)² -(x -3)² = (y-3)² -(y +6)²
(x-6-x +3)(x -6+x -3) =(y-3-y-6)(y-3+y+6)
(-3)(2x -9) = (-9)(2y +3)
2x -9 = 6y +9
2x -6y = 18
x - 3y = 9 ------(2)
from equaion (1) and (2)
2y +6(3y +9) = 14
2y + 18y +54 = 14
20y = -40
y = -2
put this in equation (2)
x = 3
hence, centre of circle =( 3, -2)
Hope this helps!
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