Show that the two spheres
$$x^2 +y^2+z^2=a^2$$
$$(x-b)^2+y^2+z^2=(b-a)^2$$
are tangent at the point $(a,0,0)$. I already obtained the gradients of the two equations which are equal to
$$\nabla f(x,y,z)= <2x, 2y, 2z>$$
$$\nabla g(x,y,z)= <2(x-b),2y,2z>$$
im having a hard time to find the constant multiplier γ please help me!!!
In the following equation system if $b=0$, the spheres are same and all points conflict. $$\begin{array}{lcl} x^2 +y^2+z^2 & = & a^2 \\ (x-b)^2 +y^2+z^2 & = & (b-a)^2 & \end{array}$$
So we can take $b \neq 0$. By subtraction of the equations $(x-b)^2 -x^2=(b-a)^2-a^2$ and we yields $2bx=2ab$. Since $b \neq 0$, $x=a$ is unique solution. We put $x=a$ in $x^2 +y^2+z^2 = a^2$ (or in $(x-b)^2 +y^2+z^2 = (b-a)^2$), we find that $y^2+z^2=0$. Hence $y=z=0$. Since $(x,y,z)=(a,0,0)$ unique solution of the system, the spheres must tangent each other.