UPSC NDA Math Model Paper 3

Let the equation of the circle be F(x,y)=x2+y2+2x +2fy+c.
Since the point (1, 0) lies on the circle, we can write F (1, 0) = 0.
Therefore,
1 + 2g + c = 0 ...(1)
Similarly, the point (0, -6) also lies on the circle therefore, F (0, -6) = 0.
Therefore,
36 - 12f + c = 0 ...(2)
Similarly, the point (3, 4) also lies on the circle therefore F (3, 4) = 0.
9 + 16 + 6g + 8f + c = 0
25 + 6g + 8f + c = 0 ...(3)
Solving the above three simultaneous equations we get:g=−

71

4

,f=

47

8

,c=

69

2

Therefore, the equation of the circle is given by:
x2+y2−2(

71

4

)x+2(

47

8

)y +

69

2

=0
x2+y2−(

71

2

)x+(

47

4

)y +

69

2

=0
4x2+4y2−142x+47y +138=0
Therefore, the equation of the circle passing through the points (1, 0), (0, -6) and (3, 4) is
4x2+4y2−142x+47y +138=0