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.
Draw a side view.
Draw a radius perpendicular to the chord, and a radius to the end of the chord.
You now have a 5-12-13 right triangle.
The circle of intersection has a radius of 12.
To find the radius of the circle formed by the intersection of a sphere and a plane, we can use the Pythagorean theorem.
First, let's find the radius of the sphere. We are given that the diameter of the sphere is 26 units. The radius of a sphere is half of its diameter, so the radius of the sphere is 26/2 = 13 units.
Next, let's find the distance between the center of the sphere and the plane. It is given that the plane is 5 units away from the center of the sphere.
Now, let's draw a straight line from the center of the sphere to the plane, perpendicular to the plane. This line is called the perpendicular distance or the height from the center of the sphere to the plane.
Using the Pythagorean theorem, we can write:
(radius of the sphere)^2 = (perpendicular distance)^2 + (distance between the center of the sphere and the plane)^2.
Substituting the values we have:
(13)^2 = (perpendicular distance)^2 + (5)^2.
Simplifying the equation:
169 = (perpendicular distance)^2 + 25.
Rearranging the equation:
(perpendicular distance)^2 = 169 - 25.
Taking the square root of both sides:
perpendicular distance = sqrt(144).
Therefore, the perpendicular distance is 12 units.
Finally, the radius of the circle formed by the intersection of the sphere and the plane is equal to the perpendicular distance, which is 12 units.