 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2017/10/c/ii/solution

Let

s = (x_{0}, y_{0}, z_{0})

be the sphere centre. The circle centre is its perpendicular projection onto the plane:

h = s - (n \cdot s - c_{0}) n, d = ∣ s - h ∣ = ∣ n \cdot s - c_{0} ∣.

For the specific plane this is

d = \frac{∣5 x _{0} + 2 y _{0} + 3 z _{0} - 11∣}{38}, h = s - \frac{5 x _{0} + 2 y _{0} + 3 z _{0} - 11}{38} (5, 2, 3) .

For a point on the section, its displacement from

h

lies in the plane and is perpendicular to

s - h

. The Pythagorean theorem therefore gives

r = R^{2} - d^{2} .

A genuine circle requires

d < R

. At

d = R

the section is a single tangent point, and at

d > R

the plane misses the sphere. A central cut

d = 0

R

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