Question

If the points $(2,0),(0,1),(4,5)$ and $(0,c)$ are concyclic, then the value of $c$ is

• A. $-1$
• B. $1$
• C. $\frac{14}{3}$
• D. $-\frac{14}{3}$

Answer 1

Answer: (B), (C)

The correct options are
B $1$
C $\frac{14}{3}$

Solution -

Let $(x,y)$ be the center of the circle

All points $(2,0),(0,1),(4,5),(0,C)$ are equidistant

from the circle

Let us find out $x,y$ by making two equations

$(x-2{)}^2+y^2=x^2+(y-1{)}^2...\left(1\right)$

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

From (1)

$4-ax=1-2y...\left(3\right)$

From (2)

$4x=37-10y...\left(4\right)$

Afters solving

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

equating the distance of different point from center

${(\frac{13}{6}-2)}^2+{\left(\frac{17}{6}\right)}^2={(\frac{13}{6}-0)}^2+{(\frac{17}{6}-c)}^2$

$\frac{1}{36}+\frac{289}{36}=\frac{169}{36}+\frac{(17-6c^2)}{36}$

$121=(17-6c{)}^2$

$17-6c=\pm 11$

$C=1,\frac{14}{3}$

Option $B,C$ are correct