Question

Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are concylic.

Answer 1

We have,
$P=(3,-2),Q=(1,0),R(-1,-2)$ and $S=(1,-4)$
let us consider $A=x^2+y^2+2gx+2fy+c=0$
Passes through P, Q and R
$\therefore 9+4+6g-4f+c=0\dots \left(ii\right)1+0+2g-0+c0\dots \left(iii\right)1+4-2g-4f+c=0\dots \left(iv\right)$
Solving (ii), (iii) and (iv) we get,
$g=-1,f=2$ and $c=1$
from (i)
The required equation of circle is
$x^2+-2x+4y+1=0\dots \left(v\right)$
Clearly s = (1, -4) satisfy (v)
Thus, P, Q, R and S are concyclic