Manipal Entrance Test 2020 Math Paper

Let equation of circle be
x2+y2+2gx+2fy+c=0
Given equation of circle is
x2+y2−4x+8=0
The centres of above circles are (−g,−f) and (2,0). Condition of orthogonality is
2(g1g2+f1f2)=c1+c2
∴‌‌2(g×(−2)+(f)×0)=c+8
⇒‌‌−4g=c+8......(i)
Also, the assume circle touch the line x+1=0.
∴ The perpendicular drawn from centre to the line is equal to radius.
∴‌‌‌

−g+1

√12

=√g2+f2−c
⇒‌‌−g+1=√g2+f2−c
g2+1−2g=g2+f2−c
⇒‌‌c=f2+2g−1
Putting the value of c in Eq. (i), we get −4g=f2+2g−1+8
⇒f2+2g+4g+7=0
⇒‌‌f2+6g+7=0
∴ Locus of centre ofcircle is y2+6x+7=0.