Question

Show that the four points $(0,0),(1,1),(5,-5)$ and $(6,-4)$ are concyclic.

Answer 1

For $4$ points $A(0,0)B(1,1)C(5,-5)$ & $D(6,-4)$ to be concyclic they should lie on the circle.

The general equation of circle is $x^2+y^2+2gx+2fy+c=0$

If A lies on circle then it should satisfy the general equation:

Hence, $0+0+0+0+c=0$,

so $c=0$

For B lies on the circle, $1+1+2g+2f+0=0$

$g+f=-1$ ……..$\left(1\right)$

For C lies on the circle, $5^2+(-5{)}^2+10g-10f+0=0$

$10g-10f=-50$ ………….$\left(2\right)$

Solving $\left(1\right)$ & $\left(2\right)$ we get $g=-3$ & $h=2$

So equation of circle is

$x^2+y^2+2(-3)x+2\left(2\right)y+0=0$

or

$x^2+y^2-6x+4y=0$ …….$\left(3\right)$

So, the fourth point should automatically satisfy this equation for the points to be concyclic.

Hence D lies on the circle.

So,

$(6{)}^2+(-4{)}^2-6\left(6\right)+4(-4)=0$

$36+16-36-16=0$

$0=0$

So, this is satisfying the equation, hence all four points are concyclic.