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| 1. |
Find the co-ordinate of center of a circle passing through the points (6,-6) , (3,-7) & (3,3) |
| Answer» Let O (x, y) is the point of the circleif 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,{tex}OA^2=OB^2=OC^2{/tex}{tex}OA^2=OB^2{/tex}{tex}=>(x-3)^2+(y+7)^2=(x-3)^2+(y-3)^2{/tex}{tex}=>(x-3)^2-(x-3)^2=(y-3)^2-(y+7)^2{/tex}{tex}=> 0=(2y+4)(3){/tex}{tex}=> y= -2{/tex}now again ,{tex}OB^2=OC^2{/tex}{tex}(x-3)^2+(y-3)^2=(x-6)^2+(y+6)^2{/tex}put y=-2{tex}=>(x-3)^2+(-2-3)^2=(x-6)^2+(-2+6)^2{/tex}{tex}=>(x-3)^2-(x-6)^2=16-25{/tex}{tex}=>(2x-9)(3)=-9{/tex}{tex}=> 2x= -3+9=6{/tex}=> x=3hence center co-ordinate is (3,-2) | |