AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Consider the equation of circle,
x2+y2+2gx+2fy+c=0........(i)
When center is (−g,−f) then it must lie on the line,
2x+3y−7=0
2g+3f+7=0.................(ii)
since equation (I) cuts the given circle so,
x2+y2−4x−6y+11=0
x2+y2−10x−4y+21=0
Orthogonally,
2g(−2)+2f(−3)=c+11
4g+6f+c+11=0.......(iii)
And,
2g(−5)+2f(2)=c+21
10g+4f+c+21=0.........(iv)
From equation (III) and (IV), 6g−2f+10=0.........(v)
From equation (II) and (V),
11f+11‌‌=0
f‌‌=−1
So,
g=−2,c=3
Then
5g−10f+3c‌‌=5(−2)−10(−1)+33
‌‌=−10+10+9
‌‌=9