question_answer

                If the circle \[{{x}^{2}}+{{y}^{2}}-6x-8y+(25-{{a}^{2}})=0\]touches the axis of \[x,\] then a equals.     JEE Main Online Paper ( Held On 23  April 2013 )

A)                  \[0\]                                    

B)                                         \[\pm 4\]

C)                                         \[\pm 2\]                                           

D)                                         \[\pm 3\]

Correct Answer: B

Solution :

                  \[{{x}^{2}}+{{y}^{2}}-6x-8y+(25-{{a}^{2}})=0\] Radius\[=4=\sqrt{9+16+(25-{{a}^{2}})}\]\[\Rightarrow \]\[a=\pm 4\]