Question

If the radius of the sphere
$x^2+y^2+z^2-2x-4y-6z$ = 0
is r, then find the value of $r^2$

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Answer 1

We want to find the radius of the sphere $x^2+y^2+z^2-2x-4y-6z$ = 0. Like we did in the case of finding centre, we will compare two equations to find the radius. For that , let the centre be (a,b,c) and the radius be r. Then the equation of the sphere will be

$(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=(r{)}^2\Rightarrow x^2+y^2+z^2-2ax-2by-2cz+a^2+b^2+c^2=r^2$

Comparing this with the given equation $x^2+y^2+z^2-2x-4y-6z=0$, we get -2 = -2a, -4 = -2b and -6 = -2c $\Rightarrow$a = 1, b = 2 and c = 3

So the centre of the sphere is (1,2,3)

We found this much in the last question.

Now, in the equation $x^2+y^2+z^2-2ax-2by-2cz+a^2+b^2+c^2=r^2$, we take all the terms to the left and compare to get radius.

$\Rightarrow x^2+y^2+z^2-2ax-2by-2cz+a^2+b^2+c^2-r^2=0......\left(1\right)$ $x^2+y^2+z^2-2x-4y-6z=0$......(2)

Comparing (1) and (2), we get

$a^2+b^2+c^2-r^2=0$
(There is no constant term in (2))

$\Rightarrow r=\sqrt{a^2+b^2+c^2}a=1,b=2andc=3\Rightarrow r=\sqrt{1^2+2^2+3^2}=\sqrt{14}\Rightarrow r^2=14$