Question

Show that the pooints (5, 5), (6, 4), (-2, 4) and (7, 1 ) all lie on a circle, and find its equation, centre and radius.

Answer 1

The general equation of circle is
$x^2+y^2+2gx+2fy+c=0\dots \left(i\right)$
centre = (-g, -f) and
radius $=\sqrt{g^2+f^2-c}$
$\because P=(5,5),Q=(6,4)$, and R = (-2, 4) lies on (i)
$\therefore 25+25+lOg+10f+c=0\dots \left(ii\right)36+16+12g+8f+c=0\dots \left(iii\right)4+16+4g+8f+c=0\dots \left(iv\right)$
Solving (ii) (iii) and (iv), we get
$g=-2,f=-1$ and $c=-20$
from (i)
The equation of circle is
$x^2+y^2-4x-2y-20=0\dots \left(A\right)$
Clearly s = (7, 1) Satisfy (A)
Hence, P, Q, R, S are concyclic
Now, centre = (-g, -f = (2, 1)
radius =$\sqrt{g^2+f^2-c}=\sqrt{4+1+20}=\sqrt{25}=5$