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 June 1st, 2017, 10:07 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 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

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