AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

Consider the given equations,
y=0 .....(I)y=x .....(II)2x+3y=10 .......(III)
Consider the diagram,The vertex A from equation (I) and (III) is (5,0).
The vertex B from equation (I) and (II) is (0,0).
The vertex C from equation (II) and (III) is (2,2).
Let the equation of circle be,
x2+y2+2gx+2fy+c=0 ....(IV)
The equation (IV) passes through point B(0,0)‌so,C=0
since, equation (IV) passes through A(5,0) so,
25+0+10g+0+0=0
g=−

5

2

The equation (IV) passes through C(2,2) so,
4+4+4g+4f=0
g+f+2=0
−

5

2

+f+2=0
f=

1

2

Therefore, the center of circle is (−g,−f)=(

5

2

,

−1

2

)