Question

1.This circle is centered at the origin and contains the points two units away. Write the equation for this circle.

The equation for this circle is x2+y2= ______


2. Show that the two points shown on the circle are solutions to your equation.
I can show that (1, 3 square root) is a solution to this equation by showing that it makes the equation true: (1) squared 2 + (3 square root) squared 2= 1 + 3 = ______ = 2 squared .

Answer 1

1. The location of the center of the given circle = The origin (0, 0)

The given points on the circumference of the circle = (1, √3), and (0, -2)

The general form of the equation of a circle, is presented as follows

(x - h)² + (y - k)² = r²

Where;

(h, k) = The coordinates of the center of the circle

r = The radius of the circle

∴ (h, k) = (0, 0)

The radius of the given equation is the distance from the center (0, 0) to either the point (0, -2) or (1, √3)

The distance from the center (0, 0) to the point (0, -2), which are points on the same ordinate, r = y₂ - y₁

∴ r = 0 - (-2) = 2

r = 2

The equation of the circle is therefore;

(x - 0)² + (y - 0)² = 2²

∴ x² + y² = 2²

2. When x = 1, and y = √3, we have;

(1 - 0)² + (√3 - 0)² = 1 + 3 = 4 = 2²

When x = 0, and y = -2, we have;

(0 - 0)² + ((-2) - 0)² = (-2)² = 4 = 2²

Therefore, the points shown on the circle (1, √3), and (0, -2) satisfy the equation of the circle, x² + y² = 2² and are solutions to the equation of the circle

Explanation: