Q120 / 160 · AP EAMCET 2011 · Mathematics
The radius of the circle given by
$\begin{aligned}
x^2+y^2 & +z^2+2 x-2 y-4 z-19=0 \\
& =x+2 y+2 z+7, \text { is }
\end{aligned}$
- A4
- Correct answerB3
- C2
- D1
Solution
Equation of the sphere,
and equation of plane
centre of the sphere $(-1,1,2)$ and radius of the sphere $\begin{aligned} R & =\sqrt{1+1+4+19} \\ & =\sqrt{25}=5 \end{aligned}$ Now, $p=$ perpendicular distance from the centre to the plane $p=\frac{|-1+2+4+7|}{\sqrt{1+4+4}}=\frac{12}{\sqrt{9}}=\frac{12}{3}=4$
$\begin{aligned} & \text { In } \Delta A O C \\ & \qquad R^2=p^2+r^2\end{aligned}$ $\begin{array}{ll}\Rightarrow & r^2=R^2-p^2 \\ \Rightarrow & r^2=25-16=9 \\ \Rightarrow & r=3\end{array}$
and equation of plane
centre of the sphere $(-1,1,2)$ and radius of the sphere $\begin{aligned} R & =\sqrt{1+1+4+19} \\ & =\sqrt{25}=5 \end{aligned}$ Now, $p=$ perpendicular distance from the centre to the plane $p=\frac{|-1+2+4+7|}{\sqrt{1+4+4}}=\frac{12}{\sqrt{9}}=\frac{12}{3}=4$
$\begin{aligned} & \text { In } \Delta A O C \\ & \qquad R^2=p^2+r^2\end{aligned}$ $\begin{array}{ll}\Rightarrow & r^2=R^2-p^2 \\ \Rightarrow & r^2=25-16=9 \\ \Rightarrow & r=3\end{array}$
Hence, the correct answer is option: 'B'.