Consider the cartesian equation of a circle, y=-2+- sqrt 36-(x-1)^2 , verses the parametric equations of the same circle: x=1+6 cos t y=-2+6sin t

Answer 1

Parametric equations:
x = 1 + 6cos(t)
y = -2 + 6sin(t)

By substituting these equations into the Cartesian equation of a circle, y = -2 + √(36 - (x - 1)^2), we can see that the parametric equations indeed represent the same circle.

y = -2 + √(36 - (1 + 6cos(t) - 1)^2)
y = -2 + √(36 - 36cos(t)^2)
y = -2 + √(36 - 36(1 - sin(t)^2))
y = -2 + √(36 - 36 + 36sin(t)^2)
y = -2 + √(36sin(t)^2)
y = -2 + 6sin(t)

Therefore, the parametric equations x = 1 + 6cos(t) and y = -2 + 6sin(t) represent the circle given by the Cartesian equation y = -2 + √(36 - (x - 1)^2).