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Question
The centre of the circle given by →r⋅(^i+2^j+2^k)=15 and ∣→r−(^j+2^k)∣=4 is-
The centre of the circle given by →r⋅(^i+2^j+2^k)=15 and ∣→r−(^j+2^k)∣=4 is-
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Solution
The correct option is B (1,3,4)
Equation of given plane and sphere in cartesian form is,
x+2y+2z−15=0=C1(say) and x2+(y−1)2+(z−2)2−4=0=C2(say)
We know intersection of a plane and a sphere will be a circle and centre of circle will lie in the plane and line joining center of sphere and circle will be normal to the plane.
Now centre of sphere is (0,1,2) and direction ratio of normal of plane are 1,2,2.
Therefore equation of line joining centres is given by,
x1=y−12=z−22=k(say)
So any point on this line is given by (k,2k+1,2k+2)
Also this point will lie in the plane:
⇒k+2(2k+1)+2(2k+2)−15=0
⇒k=1
Hence, centre of circle is (1,3,4).
Equation of given plane and sphere in cartesian form is,
x+2y+2z−15=0=C1(say) and x2+(y−1)2+(z−2)2−4=0=C2(say)
We know intersection of a plane and a sphere will be a circle and centre of circle will lie in the plane and line joining center of sphere and circle will be normal to the plane.
Now centre of sphere is (0,1,2) and direction ratio of normal of plane are 1,2,2.
Therefore equation of line joining centres is given by,
x1=y−12=z−22=k(say)
So any point on this line is given by (k,2k+1,2k+2)
Also this point will lie in the plane:
⇒k+2(2k+1)+2(2k+2)−15=0
⇒k=1
Hence, centre of circle is (1,3,4).
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the plane
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