UPSC NDA Math Model Paper 3

Let the equation of the circle be $F(x,y)=x^{2}+y^{2}+2x$ $+2 fy+c$. 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:$g=-\frac{71}{4}, f=\frac{47}{8}, c=\frac{69}{2}$ Therefore, the equation of the circle is given by: $x^{2}+y^{2}-2\left(\frac{71}{4}\right)x+2\left(\frac{47}{8}\right)y$ $+\frac{69}{2}=0$ $x^{2}+y^{2}-\left(\frac{71}{2}\right)x+\left(\frac{47}{4}\right)y$ $+\frac{69}{2}=0$ $4x^{2}+4y^{2}-142x+47y$ $+138=0$ Therefore, the equation of the circle passing through the points (1, 0), (0, -6) and (3, 4) is $4x^{2}+4y^{2}-142x+47y$ $+138=0$