A pair of perpendicular lines passes through the origin and also through the points of intersection of the curve x^2+y^2=4 with x+y=a , where a>0 . Then a is equal to

Step-by-step solution

The intersection point of the curve x^2+y^2=4 with x+y=a , where (a>0)x^2+(a-x)^2=4x^2+a^2+x^2-2 a x=42 x^2-2 a x+ (a^2-4 )=0x^2-a x+ ( a^2 2 -2 )=0x= +a a^2-4 ( a^2 2 -2 ) 2 x= a a^2-2 a^2+8 2 = a 8-a^2 2 Since, here the point of intersection should be real number, iff 8-a^2 0 ie, a^2 8, a 2 2 a 2.82 Hence, the value is a=2 according to option.