Question

Find the value of $c$ if the points $(2,0),(0,1),(4,5)$ and $(0,c)$ are concylic.

Answer 1

Concyclic point $\Rightarrow$ be on a circle ($(x,y)\equiv$ center)

then $\Rightarrow$

$(x-2{)}^2+y^2=x^2+(y-1{)}^2=(x-4{)}^2+(y-5{)}^2$

Now, solve these eqn to find $x,y,c$

$\left(1\right)\Rightarrow 4-4x=1-2y\Rightarrow 4x-2y-3=0$

$\left(2\right)\Rightarrow 1-2y=16+25-8x-10y$

$\Rightarrow 8x+8y-40=0$

$\Rightarrow x+y-5=0$

$\therefore x=\frac{13}{6}$ $y=\frac{17}{6}$

$\therefore {(\frac{13}{6}-2)}^2+{\left(\frac{17}{6}\right)}^2={\left(\frac{13}{6}\right)}^2+{\left(\frac{17-C}{6}\right)}^2$

$4-4^2\left(\frac{13}{6}\right)=C^2-2C\left(\frac{17}{6}\right)$

$12-26=3C^2-17C$

$3C^2-17C+14=0$

$3C^2-14C-3C+14=0$

$(3C-14)(C-1)=0$

$\therefore C=1$ or $C=\frac{14}{3}$