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June 1st, 2017, 10:07 PM   #1
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Sphere and Plane intersection

If $\displaystyle \beta $ is the radius of the circle of intersection of the sphere $\displaystyle x^2+y^2+z^2-2x-4y-6z+ \alpha=0$ and the plane $x+y+z=1$, then the relation between $\displaystyle \alpha $ and $\displaystyle \beta $ is :
A) $\displaystyle 3 \alpha + 3 \beta^2 =17$
B) $\displaystyle 3 \alpha^2 - 3 \beta^2 =17$
C) $\displaystyle 3 \alpha^2 - 3 \beta^2 =67$
D) $\displaystyle 3 \alpha + 3 \beta^2 =67$

My Answer is Option D.

Please check and let me know

Thank you
Lalitha183 is offline  

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intersection, plane, sphere

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