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

Correct Answer: B

Solution :

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 \[r=\frac{|2(2)-2(2)+(1)+5|}{\sqrt{4+4+1}}=\frac{|4-4+1+5|}{\sqrt{9}}\] \[r=\frac{6}{3}=2\]