The radius of the circle formed by the inters | vedclass

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(B) The equation of the sphere is ${x^2} + {y^2} + {z^2} - 2y - 4z = 11$.Rewriting it as ${x^2} + {(y - 1)^2} + {(z - 2)^2} = 11 + 1 + 4 = 16$.Thus,th

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$\sqrt{7}$

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(B) The equation of the sphere is ${x^2} + {y^2} + {z^2} - 2y - 4z = 11$.Rewriting it as ${x^2} + {(y - 1)^2} + {(z - 2)^2} = 11 + 1 + 4 = 16$.Thus,the centre of the sphere is $C = (0, 1, 2)$ and the radius $R = 4$.The distance $d$ from the centre of the sphere to the plane $x + 2y + 2z - 15 = 0$ is given by $d = \frac{|0 + 2(1) + 2(2) - 15|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|2 + 4 - 15|}{\sqrt{9}} = \frac{|-9|}{3} = 3$.The radius $r$ of the circle is given by $r = \sqrt{R^2 - d^2}$.Substituting the values,$r = \sqrt{4^2 - 3^2} = \sqrt{16 - 9} = \sqrt{7}$.

The radius of the circle formed by the intersection of the plane $x + 2y + 2z = 15$ and the sphere ${x^2} + {y^2} + {z^2} - 2y - 4z = 11$ is

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