Answer :
THE POINTS OF INTERSECTION ARE (10,0),(0,10),(-10,0) & (0,-10).
THE DISTANCE BETWEEN ANY TWO OF THESE POINTS IS
SQUARE OF DISTANCE =10^2 + 10^2
=100 + 100
=200
THEREFORE DISTANCE = √(200)
=√(5 * 5 * 2 * 2 * 2)
= 10√2
DISTANCE BETWEEN ANY TWO POINTS WILL BE 10√2 UNITS.
THE DISTANCE BETWEEN ANY TWO OF THESE POINTS IS
SQUARE OF DISTANCE =10^2 + 10^2
=100 + 100
=200
THEREFORE DISTANCE = √(200)
=√(5 * 5 * 2 * 2 * 2)
= 10√2
DISTANCE BETWEEN ANY TWO POINTS WILL BE 10√2 UNITS.
As the centre lies at (0, 0) and the radius is 10 units
The concyclic points are (10, 0), (0, 10), (0, -10) and (-10, 0)
The distance between two points = [tex] \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} [/tex]
Distance between (10,0) and (0,10) is
[tex] \sqrt{(0-10)^2 + (10-0)^2} [/tex]
[tex] \sqrt{100+ 100} = \sqrt{200} = 10 \sqrt{2} [/tex]
The concyclic points are (10, 0), (0, 10), (0, -10) and (-10, 0)
The distance between two points = [tex] \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} [/tex]
Distance between (10,0) and (0,10) is
[tex] \sqrt{(0-10)^2 + (10-0)^2} [/tex]
[tex] \sqrt{100+ 100} = \sqrt{200} = 10 \sqrt{2} [/tex]