Question

If the radius of $x^2+y^2+z^2–4x–4y+2z+K=0$is $5\mathrm{units}$. Find $K$.

Answer 1

Compute the required value.

Given: Radius of $x^2+y^2+z^2–4x–4y+2z+K=0$is $5units$.

We know that for $x^2+y^2+z^2+2ux+2vy+2wz+d=0$radius is given by:

$r=\sqrt{u^2+v^2+w^2-d}$………..…$\left(1\right)$

On comparing the given equation with the general equation, we get;

$u=-2,v=-2,w=1,d=K$

Substitute the values in equation $\left(1\right)$, we get;

$$\begin{gathered} r=\sqrt{{(-2)}^2+{(-2)}^2+{\left(1\right)}^2-K} \\ \Rightarrow 5=\sqrt{4+4+1-K} \\ \Rightarrow 5=\sqrt{9-K} \\ \Rightarrow 25=9-K[\mathrm{Squaring}\mathrm{both}\mathrm{sides}] \\ \Rightarrow K=-16 \end{gathered}$$ [ given radius is $5units$]

Hence, the value of the $K$ is $-16.$