EAMCET Engineering 2002 Solved Paper

Let the equation of the circle be
x2+y2+2gx+2fy+c=0
It passes through origin
‌ so ‌c=0
Then, the equation of circle is
x2+y2+2gx+2fy=0
It also passes through A(x1,0)
‌∴‌‌x12+0+2gx1=0
‌⇒‌‌g=−‌

x1

2

It also passes through B(0,y1)
∴o+y12+2fy1=0⇒f=‌

−y1

2

⇒ centre of the circle is (‌

x1

2

,‌

y1

2

)
Mid point of AB is (‌

x1

2

,‌

y1

2

)
i.e., OM=AM=BM
Thus, c=OM⇒c=√‌

x12

2

+‌

y12

2

x12+y12=4c2
Thus, locus is x2+y2=4c2