Answer to Question #104244 in Analytic Geometry for Deepak Rana

Question

The circle (x−1)
2 +y
2 = 1,z = 0 lies inside the sphere centred at the origin, and
having radius 2√
2.

Answer 1

To prove that the circle $\begin{cases} (x-1)^2+y^2=1 \\ z=0 \end{cases}$ lies inside sphere centred at the origin, and

having radius 2√2, we need to check that $x^2+y^2+z^2<(2\sqrt {2})^2$ .

Proof:

$\begin{cases} (x-1)^2+y^2=1 \\ z=0 \end{cases}, \quad \begin{cases} x^2-2x+1+y^2=1 \\ z=0 \end{cases}, \quad \begin{cases} x^2+y^2=2x \\ z=0 \end{cases}$

$x^2+y^2+z^2=2x+0=2x$

We know that $(x-1)^2+y^2=1 \ \Rightarrow (x-1)^2\leq1 \ \Rightarrow 0\leq x\leq 2$

So, we have $x^2+y^2+z^2=2x\leq 2\times 2=4<8=(2\sqrt{2})^2.$

Therefore, that circle lies in the sphere.