UPSC NDA I 2019 Mathematics Solved Paper

(B)

Given equation of circle is x2 + y2=a2
and equation of chord x+ y=a
For point of intersection eliminate y, we get
x2 + (a−x)2=a2
⇒ x=0, a
Hence intersection point A(0,a) and B(0, a)
Now circle with AB as diameter is
(x−0)(x−a)+( y−a)( y−0)=0
⇒ x2 + y2 − ax−ay=0