• question_answer The plane $2x-2y+z+5=0$ is a tangent to the sphere ${{(x-2)}^{2}}+{{(y-2)}^{2}}+{{(z-1)}^{2}}={{r}^{2}},$ if r equals A)  $1$                B)  $2$ C) $4$               D)  None of these

Given equation of sphere, ${{(x-2)}^{2}}+{{(y-2)}^{2}}+{{(z-1)}^{2}}={{r}^{2}}$ Here, centre of sphere $=(2,2,1)$ Let radius $=r$ Equation of a plane is,  $2x-2y+z+5=0$ Now, r = perpendicular distance on the plane from the centre of the sphere $r=\frac{|2(2)-2(2)+(1)+5|}{\sqrt{4+4+1}}=\frac{|4-4+1+5|}{\sqrt{9}}$ $r=\frac{6}{3}=2$