Show that the points $(3, - 2), (1, 0), (- 1, - 2)$ and $(1, - 4)$ are concyclic

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Solution

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$A (3, - 2), B (1, 0), C (- 1, - 2), D (1, - 4)$
$A C = \sqrt{(3 + 1)^{2} + (- 2 + 2)^{2}} = \sqrt{4^{2}} = 4$
$B D = \sqrt{(1 - 1)^{2} + (- 4)^{2}} = \sqrt{4^{2}} = 4$
$A D = \sqrt{(- 2)^{2} + (- 2)^{2}} = \sqrt{8} = 2 \sqrt{2}$
$A B = \sqrt{(- 2)^{2} + (- 2)^{2}} = 2 \sqrt{2}$
$C D = \sqrt{2^{2} + 2^{2}} = 2 \sqrt{2}$
$B C = \sqrt{(- 2)^{2} + (- 2)^{2}} = 2 \sqrt{2}$
For concyclic : $A C \times B D = (A D \times B C) + (A B \times C D)$
$4 \times 4 = (2 \sqrt{2} \times 2 \sqrt{2}) + (2 \sqrt{2} \times 2 \sqrt{2})$
$16 = 8 + 8$
$16 = 16$
$∴$ Hence, the points are concyclic

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