Is ">=1" Valid for a Circle Equation?

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Homework Statement



is:
>= 1

valid for a circle equation?
See attached?

Homework Equations



reference attached

The Attempt at a Solution



reference attached
 

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if you mean x2+y2≥1 , then yes it is a 'circle' in essence but the points (x,y) that satisfy the inequality do not lie on the circle itself like if if it was x2+y2=1
 
Hi DrMath! :smile:

(have a ≥ and try using the X2 and X2 tags just above the Reply box :wink:)

Your picture says |z - 2j| ≥ 1 …

in other words, the magnitude of the "vector" from 2j to z has length ≥ 1 …

so z is the exterior (and boundary) of the circle of length 1 and centre at 2j :smile:

(btw, it's easier to use "vectors" rather than coordinates for as problem like this :wink:)
 
Thanks tiny-tim
oh.. how to become a Homework Helper here?
 
:wink: just help people with their homework! :smile:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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