1.

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

Answer»

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Solution :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).


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