Question

If the point $(\sqrt{29},0),(5,2),(2,-5),(-1,k)$ are concylic $(k\ne 0)$ then $k$ is

• A. $\sqrt{27}$
• B. $\sqrt{28}$
• C. $\sqrt{29}$
• D. $\sqrt{26}$

Answer 1

The correct option is A $\sqrt{27}$

Points $(\sqrt{29},0),(5,2),(2,-5),(-1,k)$ are cencyclic if

$AC\times BD=AB\times CD+BC\times DA⟶\left(1\right)$

$AC=\sqrt{{(\sqrt{29}-2)}^2+{(-5-0)}^2}$

$BD=\sqrt{{(5+1)}^2+{(2-k)}^2}$

$AB=\sqrt{{(5-\sqrt{29})}^2+{(2-0)}^2}$

$BC=\sqrt{{(2-5)}^2+{(-5-2)}^2}$

$CD=\sqrt{{(2+1)}^2+{(-5-k)}^2}$

$DA=\sqrt{{(-1-\sqrt{29})}^2+{\left(k\right)}^2}$

putting these values in equation $\left(1\right)$ we get

$K=\sqrt{27}$