ok if i have a hyperbolic line say |z-6|=3 this means centre at (6,0) and radius 3 correct? when i turn it into terms with x and y does it go to.... (x-6)^2 +y^2 = 3 ORRRR (x-6)^2 +y^2 = 9 because take the square root over?
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Originally Posted by skystar ok if i have a hyperbolic line say |z-6|=3 this means centre at (6,0) and radius 3 correct? when i turn it into terms with x and y does it go to.... (x-6)^2 +y^2 = 3 ORRRR (x-6)^2 +y^2 = 9 because take the square root over? It's $\displaystyle \sqrt{(x - 6)^2 + y^2} = 3$ so the second expression. -Dan
Originally Posted by skystar ok if i have a hyperbolic line say |z-6|=3 this means centre at (6,0) and radius 3 correct? when i turn it into terms with x and y does it go to.... (x-6)^2 +y^2 = 3 ORRRR (x-6)^2 +y^2 = 9 because take the square root over? Geometrically |z-6|=3 is the set of points in the complex plane that are a constant distance 3 units from the point z = 6 ......
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