Question

The angle of intersection of the two spheres \(x^2+y^2+z^2+6y+2z+8=0\) and \(x^2+y^2+z^2+6x+8y+4z+20=0\) is

• A. \(\dfrac{\pi}{2}\)
• B. \(\dfrac{\pi}{3}\)
• C. \(\dfrac{\pi}{6}\)
• D. \(\dfrac{\pi}{4}\)

Hint:

For angle of intersection of two spheres, use \(\cos\theta=\frac{r_1^2+r_2^2-d^2}{2r_1r_2}\).

Answer 1

The Correct Option is A

Concept:
The angle of intersection of two spheres is the angle between their tangent planes at a common point. It is also the angle between their radii drawn to that common point. If the centres are separated by distance \(d\), and radii are \(r_1,r_2\), then \[ \cos\theta=\frac{r_1^2+r_2^2-d^2}{2r_1r_2} \]

Step 1: Find centre and radius of first sphere. First sphere: \[ x^2+y^2+z^2+6y+2z+8=0 \] Complete squares: \[ y^2+6y=(y+3)^2-9 \] \[ z^2+2z=(z+1)^2-1 \] So, \[ x^2+(y+3)^2-9+(z+1)^2-1+8=0 \] \[ x^2+(y+3)^2+(z+1)^2=2 \] Therefore, \[ C_1=(0,-3,-1) \] and \[ r_1=\sqrt2 \]

Step 2: Find centre and radius of second sphere. Second sphere: \[ x^2+y^2+z^2+6x+8y+4z+20=0 \] Complete squares: \[ x^2+6x=(x+3)^2-9 \] \[ y^2+8y=(y+4)^2-16 \] \[ z^2+4z=(z+2)^2-4 \] So, \[ (x+3)^2-9+(y+4)^2-16+(z+2)^2-4+20=0 \] \[ (x+3)^2+(y+4)^2+(z+2)^2=9 \] Therefore, \[ C_2=(-3,-4,-2) \] and \[ r_2=3 \]

Step 3: Find distance between centres. \[ d=\sqrt{(-3-0)^2+(-4+3)^2+(-2+1)^2} \] \[ d=\sqrt{(-3)^2+(-1)^2+(-1)^2} \] \[ d=\sqrt{9+1+1} \] \[ d=\sqrt{11} \]

Step 4: Apply formula. \[ \cos\theta=\frac{r_1^2+r_2^2-d^2}{2r_1r_2} \] \[ \cos\theta=\frac{2+9-11}{2(\sqrt2)(3)} \] \[ \cos\theta=\frac{0}{6\sqrt2} \] \[ \cos\theta=0 \] So, \[ \theta=\frac{\pi}{2} \]

Step 5: Final answer. \[ \boxed{\frac{\pi}{2}} \]