If $\lambda x^2+5y^2+\mu z^2-axy+2\lambda x-4\mu y-2az+d=0$ represents a sphere of radius $2.$ Find the value of $d$

Question

If $\lambda x^2+5y^2+\mu z^2-axy+2\lambda x-4\mu y-2az+d=0$ represents a sphere of radius $2.$ Find the value of $d.$

My solution goes like this:

Since, $\lambda x^2+5y^2+\mu z^2-axy+2\lambda x-4\mu y-2az+d=0$ is an equation of a sphere, so, $\lambda=\mu=5$ and $a=0.$ Hence, $5x^2+5y^2+5z^2+10x-20y+d=0\implies x^2+y^2+z^2+2x-4y+\frac d5=0.$ Given, $\text{radius}(r)=2$ and hence $$\sqrt{1+4+0-\frac d5}=2\implies d=5.$$

Is the above solution correct? If not, where is it going wrong?

Answer 1

You are indeed correct.

I would have added a couple steps between

$x^2+y^2+z^2+2x-4y+(d/5)=0$

and the formula you used to compute the radius. To wit, combine terms with like variables and complete the squares:

$(x^2+2x)+(y^2-4y)+z^2=-(d/5)$

$(x+1)^2+(y-2)^2+z^2=1+4-(d/5)$

making it easier to see where the formula for the radius comes from.