pre cal prob
Remember the distance formula, based on the pythagorean theorem?
If you have a horizontal distance x and a vertical distance y, then the hypotenuse will be x² + y²
So your formula is x² + y² = 2²
or
x² + y² = 4
Or solving in terms of y:
y² = 4 – x²
y = sqrt(4 – x²)
pythagoras’ theorem
c^2 = a^2 + b^2
4 = x^2 + y^2
x^2 + y^2 – 4 = 0
y^2 = 4 – x^2
y = sqrt(4 – x^2)
x ² + y ² = 2 ²
x ² + y ² = 4
(the equation of a circle centre O , radius 2)