Answer to Question #100457 in Analytic Geometry for Akash

Question

Find the radius of the circular section of the
sphere Ir—cl = 7 by the plane
r. (3i — j + 2k) = 2 ,r7 , where c = (— 1, 0, 1).

Answer 1

Let 𝑆 be the sphere in 𝑅³ with center 𝑂(𝑥₀, 𝑦₀, 𝑧₀) and radius 𝑟, and let 𝑃 be the plane with equation 𝐴𝑥 + 𝐵𝑦 + 𝐶𝑧 = 𝐷,

so that 𝑛⃗ = (𝐴, 𝐵, 𝐶) is a normal vector of 𝑃.

If 𝑃₀ is an arbitrary point on 𝑃, the signed distance from the center of the sphere 𝑂 to the plane 𝑃 is

𝜌 = (𝑂 − 𝑃₀)𝑛⃗/ |𝑛⃗|

= (𝐴𝑥₀ + 𝐵𝑦₀ + 𝐶𝑧₀ − 𝐷)/(A²+B²+C²)^1/2

The intersection 𝑆 ∩ 𝑃 is a circle if and only if −𝑟 < 𝜌 < 𝑟, and in that case, the circle has radius

𝑟_c = $\sqrt{r^2-\rho^2}$  and

center 𝑐 = 𝑂 + 𝜌 ∙ 𝑛⃗/ |𝑛⃗ |

=(𝑥₀, 𝑦₀, 𝑧₀) + 𝜌 ∙ (𝐴, 𝐵, 𝐶)/(A²+B²+C² )^1/2

so in our case,

|𝑟| =7 and 𝑂(𝑥₀, 𝑦₀, 𝑧₀)=(-1,0,1) and P={3x - y + 2z = 2}

𝜌= (𝐴𝑥₀ + 𝐵y₀ + 𝐶𝑧₀ −D)/(A²+B²+C²)^1/2

= (3.(-1)+0+1.2-2)/(14)^1/2

= -3/(14)^1/2

=-0.80

𝑟_c = (𝑟² − 𝜌² )^1/2 = (49-0.64)^1/2 =6.95,

hence the radius of this circular plane is 6.95.