Question

The condition that the two spheres $a(x^2+y^2+z^2)+2lx+2my+2nz+p=0$ and $b(x^2+y^2+z^2)=k^2$ may cut orthogonally $(k\ne 0)$

• A. $bp=a{k}^2$
• B. $b{k}^2=ap$
• C. $\frac{k^2}{p}=\frac{-p}{a}$
• D. $a=p^{2}bk$

Answer 1

The correct option is A $bp=a{k}^2$

First sphere is

$a(x^2+y^2+z^2)+2lx+2my+2nz+p=0$

$x^2+y^2+z^2+2\left(\frac{l}{a}\right)x+2\left(\frac{m}{a}\right)y+2\left(\frac{n}{a}\right)z+\frac{p}{a}=0$ ....(1)

Second sphere is:

$b(x^2+y^2+z^2)=k^2$

$\Rightarrow x^2+y^2+z^2+2\left(0\right)x+2\left(0\right)y+2\left(0\right)z-\frac{k^2}{b}=0$ .....(2)

Applying condition of orthogonality

$2u_1{u}_2+2v_1{v}_2+2w_1{w}_2=d_1+d_2$, we get

$0=\left(\frac{p}{a}\right)-\left(\frac{k^2}{b}\right)$

$\therefore a{k}^2=bp$