Question

Find the radius of the circle formed by the intersection of a sphere of diameter 26 units and a plane that is 5 units away from the center of the sphere.

Answer 1

Answer:

12 units

Explanation:

This problem can be solved if we take into account the equation for a sphere

$x^{2}+y^{2}+z^{2}=r^{2}\\x^{2}+y^{2}+z^{2}=(\frac{26}{2})^{2}=13^{2}$

where we took that the radius is 13 units. If we take z=5 and we replace this value in the equation of the sphere we have

$x^{2}+y^{2}+(5)^{5}=13^{2}\\x^{2}+y^{2}=144=(12)^{2}$

where we have taken x2 +y2 because if the equation of a circunference.

In this case the intersection is made when we take z=5, for this value the sphere and the plane coincides in values.

Hence, the radius is 12 units

I hope this is useful for you

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