Two points from the set of concyclic points o | vedclass

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(B) The equation of the circle passing through $(1,1), (2,-1),$ and $(3,2)$ is given by the determinant equation: $\left|\begin{array}{cccc} x^2+y^2 &

Vedclass · Accepted Answer

$\left(\frac{5}{2}+\sqrt{\frac{5}{2}}, \frac{1}{2}\right), \left(\frac{5+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$

Vedclass · Answer

(B) The equation of the circle passing through $(1,1), (2,-1),$ and $(3,2)$ is given by the determinant equation: $\left|\begin{array}{cccc} x^2+y^2 & x & y & 1 \ 2 & 1 & 1 & 1 \ 5 & 2 & -1 & 1 \ 13 & 3 & 2 & 1 \end{array}ight| = 0$ Simplifying the determinant,we get: $x^2+y^2-5x-y+4 = 0$ Testing the points in option $B$: For $\left(\frac{5+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}ight)$: $\left(\frac{5+\sqrt{5}}{2}ight)^2 + \left(\frac{1+\sqrt{5}}{2}ight)^2 - 5\left(\frac{5+\sqrt{5}}{2}ight) - \left(\frac{1+\sqrt{5}}{2}ight) + 4 = 0$ $\frac{25+5+10\sqrt{5}}{4} + \frac{1+5+2\sqrt{5}}{4} - \frac{25+5\sqrt{5}}{2} - \frac{1+\sqrt{5}}{2} + 4 = 0$ $\frac{36+12\sqrt{5}}{4} - \frac{26+6\sqrt{5}}{2} + 4 = 9+3\sqrt{5} - 13-3\sqrt{5} + 4 = 0$ Thus,the points in option $B$ satisfy the circle equation.

Two points from the set of concyclic points of the circle passing through $(1,1), (2,-1),$ and $(3,2)$ are:

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