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

Options

A. `(3,2)`
B. `(3,-2)`
C. `(3,-3)`
D. None

Answer

B

Explanation

Let C(x, y) be the centre of the circle passing through the points
P(6, -6), Q(3, -7) and R(3, 3).

Then, `" "PC=QC=CR" "` (radius of circle)
Now, `" "PC=QC`
`rArr" "PC^(2)=QC^(2)`
`rArr" "(x-6)^(2)+(y+6)^(2)=(x-3)^(2)+(y+7)^(2)`
`" "[because "distance "=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))]`
`rArr" "x^(2)-12x+36+y^(2)+12y+36=x^(2)-6x+9+y^(2)+14y+49`
`rArr" "-12x+6x+12y-14y+72-58=0" "rArr" "-6x-2y+14=0`
`" "3x+y-7=0" "("divide by"-2 )...(1)`
and `" "QC=CR`
`rArr" "QC^(2)=CR^(2)`
`rArr" "(x-3)^(2)+(y+7)^(2)=(x-3)^(2)+(y-3)^(2)`
`rArr" "x^(2)-6x+9+y^(2)+14y+49=x^(2)-6x+9+y^(2)-6y+9`
`rArr" "-6x+6x+14y+6y+58-18=0`
`rArr" "20y+40=0" "rArr" "y=-(40)/(20)=-2" "...(2)`
Putting y=-2 in Eq. (1) , we get
`" "3x-2-7=0`
`rArr" "3x=9" "rArr" "x=3`
Hence, centre is (3, -2).