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
$\begin{vmatrix}x^{2}+y^{2} & x & y & 1 \\ 1+1 & 1 & 1 & 1 \\ 4+1 & 2 & -1 & 1 \\ 9+4 & 3 & 2 & 1\end{vmatrix}=0$
$\begin{vmatrix}x^{2}+y^{2} & x & y & 1 \\ 2 & 1 & 1 & 1 \\ 5 & 2 & -1 & 1 \\ 13 & 3 & 2 & 1\end{vmatrix}=0$
$\begin{vmatrix}x^{2}+y^{2}-2 & x-1 & y-1 & 0 \\ -3 & -1 & 2 & 0 \\ -8 & -1 & -3 & 0 \\ 13 & 3 & 2 & 1\end{vmatrix}=0$
$\begin{vmatrix}x^{2}+y^{2}-2 & x-1 & y-1 \\ -3 & -1 & 2 \\ -8 & -1 & -3\end{vmatrix}=0$
$\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 $
$x^{2}+y^{2}-2-5 x+5-y+1=0$
$x^{2}+y^{2}-5 x-y+4=0$
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.