Question

Radius of the circle \( \vec{r}^2 + \vec{r} \cdot (2\hat{i} - 2\hat{j} - 4\hat{k}) - 19 = 0, \vec{r} \cdot (\hat{i} - 2\hat{j} + 2\hat{k}) + 8 = 0 \) is

• A. 5
• B. 4
• C. 3
• D. 2

Hint:

Radius of the circle $\vecr+\vecr·(2\hati-2\hatj-4\hatk)-19=0, \vecr·(\hati-2\hatj+2\hatk)+8=0$ is

Answer 1

The Correct Option is B

Step 1: Concept
A circle in 3D is the intersection of a sphere and a plane.
Step 2: Analysis
Sphere: $x^2+y^2+z^2+2x-2y-4z-19=0$ with center $C(-1, 1, 2)$ and radius $R = \sqrt{1^2+(-1)^2+(-2)^2-(-19)} = 5$.
Step 3: Evaluation
Perpendicular distance $P$ from $C(-1, 1, 2)$ to plane $x-2y+2z+8=0$ is $P = \frac{|-1-2(1)+2(2)+8|}{\sqrt{1^2+(-2)^2+2^2}} = \frac{9}{3} = 3$.
Step 4: Conclusion
Radius of circle $= \sqrt{R^2 - P^2} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4$.
Final Answer: (b)