Drawing Graphs: Concentric Circles & Straight Lines

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To draw concentric circles centered at (0,0), the equation ln(x^2 + y^2) = constant can be transformed into x^2 + y^2 = e^constant, resulting in circles with radius √e^constant. For straight lines through (0,0), the equation arctan(y/x) = constant leads to y = (tan(constant))x, indicating lines with a slope of tan(constant). The discussion clarifies that ln(0) is undefined, and x=0 does not yield valid solutions for these equations. The explanations provided enhance understanding of the relationships between constants and their graphical representations. Overall, the thread effectively addresses how to graph these functions and their implications.
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Hi, how would I go about drawing these two graphs?

31e4a231b2f01490ad29b8db02cbfd4c.png


and

3bbb0c98927abe3e79236b39a7587ef4.png
The first one would be concentric circles with the centre at (0,0).
The second one would be straight lines through (0,0).

Is this correct?
Also, what happens at ln(0) = constant for the first graph and x = 0 for the second graph?

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Correct.
Raees said:
Also, what happens at ln(0) = constant for the first graph
That is not possible. x=y=0 is never a solution. Same for x=0 in the second case.

For "interesting" functions f(x)=constant usually means x has one or a somewhat special set of values for a given constant. If f is injective then it is equivalent to x=constant.
 
For the first one, ln(x^2+ y^2)= constant, taking the exponential of both sides, x^2+ y^2= e^{constant}. Yes, the graph of that is a circle with center at the origin and radius \sqrt{e^{constant}}.

For the second one, arctan(y/x)= constant, taking the tangent of both sides, y/x= tan(constant) so y= (tan(constant))x. Since the tangent of a constant is also a constant, this is a straight line through the origin with slope tan(constant).
 
HallsofIvy said:
For the first one, ln(x^2+ y^2)= constant, taking the exponential of both sides, x^2+ y^2= e^{constant}. Yes, the graph of that is a circle with center at the origin and radius \sqrt{e^{constant}}.

For the second one, arctan(y/x)= constant, taking the tangent of both sides, y/x= tan(constant) so y= (tan(constant))x. Since the tangent of a constant is also a constant, this is a straight line through the origin with slope tan(constant).

Thanks, that helps a lot!
 
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