The radius of the circle given by $\begin{aligned} x^2+y^2 & +z^2+2 x-2 y-4 z-19=0 \ & =x+2 y+2 z+7, ext { is } \end{aligned}$

Question

The radius of the circle given by $\begin{aligned} x^2+y^2 & +z^2+2 x-2 y-4 z-19=0 \ & =x+2 y+2 z+7, 	ext { is } \end{aligned}$

Aicharya · Accepted Answer

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 $p=\frac{|-1+2+4+7|}{\sqrt{1+4+4}}=\frac{12}{\sqrt{9}}=\frac{12}{3}=4$

$\begin{aligned} & \text { In } \Delta A O C \\ & \qquad R^2=p^2+r^2\end{aligned}$ $\begin{array}{ll}\Rightarrow & r^2=R^2-p^2 \\ \Rightarrow & r^2=25-16=9 \\ \Rightarrow & r=3\end{array}$

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…

Solution