KEAM 2012 Math Paper

since, P(−2,2) and Q(k,0) is the end points of a diameter.
∴ Mid-point M of PQ=‌

−2+k

2

,‌

2+0

2

i.e., M=(‌

k

2

−1,1)
Also, MP= radius of circle
⇒‌‌√(−1−‌

k

2

)2+1=1
⇒‌‌1+‌

k2

4

+k+1=1
⇒‌‌k2+4k+4=0
⇒‌‌(k+2)2=0
⇒ k=−2
Mid-point M=(−‌

2

2

−1,1)=(−2,1)
which is equal to the centre of circle.
∴ Equation of circle is
(x+2)2+(y−1)2=12
⇒‌‌x2+4x+4+y2−2y+1=1
⇒‌‌x2+y2+4x−2y+4=0