Question

Show that the four points (1,2) (3-4) (5-6) (19,8)

Answer 1

Let A(1,2), B(3,-4), C(5,-6), D(19,8) be the points on the circle. For the points to be concyclic, OA=OB=OC=OD=r, the radius of the circle.

For chord-AB, midpoint is (2,-1). Slope(AB)=-6/2=-3. Perpendicular bisector of a AB passes through the centre of circle. Slope of perpendicular bisector of AB=-1/-3=1/3. equation of perpendicular bisector of AB is y=x/3+c1 => x-3y+c1'=0. As it passes through the said midpoint, 2+3+c1'=0 => c1'=-5. Thus x-3y-5=0 is the line passing through the centre of the circle.

For chord-BC, midpoint is (4,-5). Slope(BC)=-2/2=-1. Perpendicular bisector of a chord passes through the centre of circle. Slope of perpendicular bisector of BC=-1/-1=1. equation of perpendicular bisector of BC is y=x+c2 => x-y+c2=0. As it passes through the said midpoint 4+5+c2=0 => c2=-9. Thus x-y-9=0 is a line which is also passing through the centre of the circle.

Solving equations x-3y-5=0 & x-y-9=0, gives O(11,2) as centre of the circle. The equation of the required circle is (x-11)^2+(y-2)^2=r^2.

At A(1,2), OA^2=(1-11)^2+(2-2)^2=100 (radius=10 units)

At B(3,-4), OB^2=(3-11)^2+(-4-2)^2=64+36=100

At C(5,-6), OC^2=(5-11)^2+(-6-2)^2=36+64=100

At D(19,8), OD^2=(19-11)^2+(8-2)^2=64+36=100

Thus, the given coordinates are concyclic on (lie on the circle) (x-11)^2+(y-2)^2=100