**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*

**Answer:**

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

**Answer:**

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

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