Question

Two points from the set of Concyclic points of the circle passing through $(1,1),(2,-1),(3,2)$ is

• A. $\left(\frac{5}{2}+\sqrt{\frac{5}{2}},\frac{1}{2}+\sqrt{\frac{5}{2}}\right),\left(\frac{5}{2},\frac{1}{2}+\sqrt{\frac{5}{2}}\right)$
• B. $\left(\frac{5}{2}+\sqrt{\frac{5}{2}},\frac{1}{2}\right),\left(\frac{5+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}\right)$
• C. $\left(\frac{5+\sqrt{5}}{2},\frac{1+\sqrt{5}}{\sqrt{2}}\right),\left(\frac{5}{2}+\sqrt{\frac{5}{2}},\frac{1+\sqrt{5}}{4}\right)$
• D. $\left(\frac{5}{2}-\frac{\sqrt{5}}{2^{\prime }}\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(\frac{5}{2}-\frac{\sqrt{5}}{2},\frac{1}{2}+\frac{\sqrt{5}}{2}\right)$

Answer 1

Answer: (B)

Equation of circle passing through the points $(1,1),(2,-1)$ and $(3,2)$ is
$\left|\begin{array}{cccc}{x}^2+y^2& x& y& 1\\ 1+1& 1& 1& 1\\ 4+1& 2& -1& 1\\ 9+4& 3& 2& 1\end{array}\right|=0$
$\left|\begin{array}{cccc}{x}^2+y^2& x& y& 1\\ 2& 1& 1& 1\\ 5& 2& -1& 1\\ 13& 3& 2& 1\end{array}\right|=0$
$\left|\begin{array}{cccc}{x}^2+y^2-2& x-1& y-1& 0\\ -3& -1& 2& 0\\ -8& -1& -3& 0\\ 13& 3& 2& 1\end{array}\right|=0$
$\left|\begin{array}{ccc}{x}^2+y^2-2& x-1& y-1\\ -3& -1& 2\\ -8& -1& -3\end{array}\right|=0$
$\left(x^2+y^2-2\right)(3+2)-(x-1)(9+16)+(y-1)(3-8)=0$
$5\left(x^2+y^2-2\right)-25(x-1)-5(y-1)=0$
$x^2+y^2-2-5(x-1)-(y-1)=0$
$x^2+y^2-2-5x+5-y+1=0$
$x^2+y^2-5x-y+4=0$
Now, by putting all the points, we see
$\left(\frac{5}{2}+\sqrt{\frac{5}{2}},\frac{1}{2}\right\}\left(\frac{5+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}\right)$
passes through above circle.