The radius of the sphere x^2+y^2+z^2=12x+4y+3 | vedclass

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(A) The given equation of the sphere is $x^2+y^2+z^2-12x-4y-3z=0$. Comparing this with the general equation of a sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$,

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$\frac{13}{2}$

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(A) The given equation of the sphere is $x^2+y^2+z^2-12x-4y-3z=0$. Comparing this with the general equation of a sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$, we get $2u=-12$, $2v=-4$, and $2w=-3$. Thus, $u=-6$, $v=-2$, and $w=-\frac{3}{2}$. The centre of the sphere is $(-u, -v, -w) = (6, 2, \frac{3}{2})$. The radius $r$ of the sphere is given by the formula $r = \sqrt{u^2+v^2+w^2-d}$. Substituting the values, we get $r = \sqrt{(-6)^2+(-2)^2+(-\frac{3}{2})^2-0}$. $r = \sqrt{36+4+\frac{9}{4}} = \sqrt{40+\frac{9}{4}} = \sqrt{\frac{160+9}{4}} = \sqrt{\frac{169}{4}}$. Therefore, $r = \frac{13}{2}$.

The radius of the sphere $x^2+y^2+z^2=12x+4y+3z$ is

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