Question

If $(2,0),(0,1),(4,5)$ and $(0,C)$ are concyclics then find $C$

Answer 1

Let $A=(2,0),B=(0,1),C=(4,5),D=(0,C)$

If $\square ABCD$ is a cyclic quadrilateral, product of diagonals has to be equal to the sum of the product of the opposite sides.

$\therefore AC.BD=AB.CD+BC.AD$

$\therefore \sqrt{(2-4{)}^2+(0-5{)}^2}.\sqrt{0+(1-C{)}^2}=$ $\sqrt{(2-0{)}^2+(0-1{)}^2}.\sqrt{(4-0{)}^2+(5-C{)}^2}+\sqrt{(0-4{)}^2+(1-5{)}^2}.\sqrt{(2-0{)}^2+(0-C{)}^2}$

Squaring both sides, we have

$29\times (C-1{)}^2=5\times (16+25+C^2-10C)+2\sqrt{\left(5\right)(16+25-10C+C^2)\left(32\right)(C^2+4)}+32\times (C^2+4)$

$\therefore 29C^2-58C+29-32C^2-128-205-5C^2+50C=2\sqrt{\left(160\right)(C^2+4)(41-10C+C^2)}$

$\therefore -8C^2-8C-304=2\sqrt{160(C^2+4)(41-10C+C^2)}$

$\therefore -(C^2+C+38)=\sqrt{10(C^2+4)(41-10C+C^2)}$

Squaring both sides, we have

$C^4+C^2+76C+1444+2C^3+76C^2=10(C^4-10C^3+41C^2+164-40C+4C^2)$

$\therefore 9C^4-102C^3+373C^2-476C+196=0$

By observation, we can conclude that $C=1$

$\therefore (C-1)(9C^3-93C^2+280C-196)=0$

Again, $(C-1{)}^2(9C^2-84C+196)=0$

$\therefore (C-1{)}^2(3C-14{)}^2=0$

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