question_answer

If the equation of one tangent to the circle with centre at (2, ?1) from the origin is \[3x+y=0\], then the equation of the other tangent through the origin is

A)            \[3x-y=0\]                                

B)            \[x+3y=0\]

C)            \[x-3y=0\]                                

D)            \[x+2y=0\]

Correct Answer: C

Solution :

           Centre is \[(2,\ -1)\]. Therefore \[r=\left| \frac{3(2)-1}{\sqrt{10}} \right|\ =\frac{5}{\sqrt{10}}\]                     Now draw a perpendicular on \[x-3y=0\], we get                    \[r=\left| \frac{2-3(-1)}{\sqrt{10}} \right|\ =\frac{5}{\sqrt{10}}\].