Question

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

Please answer it fast , very urgent!!!!!!

Answer 1

For the point (0,0),(1,1),(5.-5),(6,-4) to be concyclic They should lie on a circle
It should satisfy (x-a)² + (y-b)²=r² where (a,b) is the centre of circle and r is the radius of circle
so substituting (0,0) a²+b²=r² .....say eq1
Substituting (1,1) we get a² +1-2a+b²+1-2b=r²
2-2(a+b)=0     (∵eq1) ;So, a+b=1 .... say eq2
Substituting (5,-5) we get a² + 25 -10a +b² +25 +10b=r²
50+10(b-a)=0  (∵eq1)
a-b=5 .... say eq3
eq3 + eq2 we get a-b+a+b=5+1
2a=6 ; a=3
substituting a=3 in eq 2 we get b=-2
Substituting both vaues in eq1 we get
r=√13

If (6,-4) lies on this circle it should satisfy (x-3)²+(y+2)²=13
So,(6-3)² + (-4+2)²=3²+2²=13
i.e (6.-4) lies on Same cirlce

So all the given point are Concyclic to a Circle whose centre is (3,-2)
and Radius of √13