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# If the radius of the sphere x2+y2+z2−2x−4y−6z = 0 is r, then find the value of r2 ___

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## We want to find the radius of the sphere x2+y2+z2−2x−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⇒x2+y2+z2−2ax−2by−2cz+a2+b2+c2=r2 Comparing this with the given equation x2+y2+z2−2x−4y−6z=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+z2−2ax−2by−2cz+a2+b2+c2=r2, we take all the terms to the left and compare to get radius. ⇒x2+y2+z2−2ax−2by−2cz+a2+b2+c2−r2=0......(1) x2+y2+z2−2x−4y−6z=0......(2) Comparing (1) and (2), we get a2+b2+c2−r2=0(There is no constant term in (2) ⇒r=√a2+b2+c2a=1, b=2 and c=3⇒r=√12+22+32=√14⇒r2=14  Suggest Corrections  0      Similar questions  Explore more