If the angle between the pair of straight lines formed by joining the points of intersection of x^2+y^2=4 and y=3 x+c to the origin is a right angle, then c^2 is

Step-by-step solution

aligned & x^2+y^2=4 x^2+y^2-4(1)^2=0 ...(i) & and y=3 x+c y-3 x c =1 aligned Put the value of ' 1 ' in Eq. (i) aligned & x^2+y^2-4 ( y-3 x c )^2=0 & x^2+y^2-4 ( ( y c )^2+ ( 3 x c )^2- 6 x y c^2 )=0 & x^2 (1- 36 c^2 )+y^2 (1- 4 c^2 )+ 24 x y c^2 =0 aligned According to the question, aligned & (1- 36 c^2 )+ (1- 4 c^2 )=0 2= 40 c^2 & c^2=20 aligned