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Question 1. Asked on :23 September 2019:09:44:14 AM

#### find the center of circle passing though the points (6,-6),(3,-7)and (3,3).

-Added by Rohit Rajput Mathematics » Cooridinate Triangle

Akki chauhan

let O (x, y) is the point of circle

if three given points A (3,-7) B (3,3) and C (6,-6)

we know distance between circumference and center is always same. i.e radius .
now,
OA^2=OB^2=OC^2

OA^2=OB^2
=>(x-3)^2+(y+7)^2=(x-3)^2+(y-3)^2

=>(x-3)^2-(x-3)^2=(y-3)^2-(y+7)^2

=> 0=(2y+4)(3)

=> y= -2
now again ,
OB^2=OC^2
(x-3)^2+(y-3)^2=(x-6)^2+(y+6)^2
put y=-2

=>(x-3)^2+(-2-3)^2=(x-6)^2+(-2+6)^2

=>(x-3)^2-(x-6)^2=16-25

=>(2x-9)(3)=-9

=> 2x= -3+9=6

=> x=3

hence center co-ordinate is (3,-2)

-Answered by Akki chauhan On 23 September 2019:08:50:58 PM(10905Average Rating Based on rating)

priyanshu kumar

let O (x, y) is the point of circle

if three given points A (3,-7) B (3,3) and C (6,-6)

we know distance between circumference and center is always same. i.e radius .
now,
OA^2=OB^2=OC^2

OA^2=OB^2
=>(x-3)^2+(y+7)^2=(x-3)^2+(y-3)^2

=>(x-3)^2-(x-3)^2=(y-3)^2-(y+7)^2

=> 0=(2y+4)(3)

=> y= -2
now again ,
OB^2=OC^2
(x-3)^2+(y-3)^2=(x-6)^2+(y+6)^2
put y=-2

=>(x-3)^2+(-2-3)^2=(x-6)^2+(-2+6)^2

=>(x-3)^2-(x-6)^2=16-25

=>(2x-9)(3)=-9

=> 2x= -3+9=6

=> x=3

hence center co-ordinate is (3,-2)

-Answered by priyanshu kumar On 02 October 2019:12:31:07 PM(11530Average Rating Based on rating)

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