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Question

If the radius of the sphere x2+y2+z22x4y6z = 0 is r, then find the value of r2


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Solution

We want to find the radius of the sphere x2+y2+z22x4y6z = 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

(xa)2+(yb)2+(zc)2=(r)2x2+y2+z22ax2by2cz+a2+b2+c2=r2

Comparing this with the given equation x2+y2+z22x4y6z=0, we get -2 = -2a, -4 = -2b and -6 = -2c 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 x2+y2+z22ax2by2cz+a2+b2+c2=r2, we take all the terms to the left and compare to get radius.

x2+y2+z22ax2by2cz+a2+b2+c2r2=0......(1) x2+y2+z22x4y6z=0......(2)

Comparing (1) and (2), we get

a2+b2+c2r2=0(There is no constant term in (2)

r=a2+b2+c2a=1, b=2 and c=3r=12+22+32=14r2=14

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