Question

Radius of the circle for sphere ${\overrightarrow{r}}^2+\overrightarrow{r}.(2\overrightarrow{i}-2\overrightarrow{j}-4\overrightarrow{k})-19=0$ and plane $\overrightarrow{r}.(\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k})+8=0$

• A. $5$
• B. $4$
• C. $3$
• D. $2$

Answer 1

The correct option is B $4$

Given circle is intersection of sphere

$x^2+y^2+z^2+2x-2y-4z-19=0$ ...(1)

and plane $x-2y+2z+8=0$ ...(2)

Center of the sphere is $(-1,1,2)$

$p=$ length of the $\perp$ from, $(-1,1,2)$ upon (2)

$=\frac{-1-2+4+8}{\sqrt{1+4+4}}=\frac{9}{3}=3$

$R=$ radius of the sphere $=\sqrt{1+1+4+19}=5$

Radius of the circle $=\sqrt{R^2-p^2}=\sqrt{25-9}=\sqrt{16}=4$