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Question:
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The radius of the circle given by
$\begin{aligned}
x^2+y^2 & +z^2+2 x-2 y-4 z-19=0 \\
& =x+2 y+2 z+7, \text { is }
\end{aligned}$
Options:
$\begin{aligned}
x^2+y^2 & +z^2+2 x-2 y-4 z-19=0 \\
& =x+2 y+2 z+7, \text { is }
\end{aligned}$
Solution:
2333 Upvotes
Verified Answer
The correct answer is:
3
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
$p=\frac{|-1+2+4+7|}{\sqrt{1+4+4}}=\frac{12}{\sqrt{9}}=\frac{12}{3}=4$
$\begin{aligned} & \text { In } \Delta A O C \\ & \qquad R^2=p^2+r^2\end{aligned}$
$\begin{array}{ll}\Rightarrow & r^2=R^2-p^2 \\ \Rightarrow & r^2=25-16=9 \\ \Rightarrow & r=3\end{array}$
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