Does an Ellipse Intersecting a Circle Result in Imaginary Numbers?

[femas]

Hi

Let's say that we have equation of circle as

x² + y² = R²

and equation of ellipse in quadratic form as

A x² + B y² + Cx + D = 0

if the circle is inside the ellipse, so there is no intersection ...
Are x and y imag in this case? /or/ is one of them imag and the other is real ?... etc.

 

[rock.freak667]

No intersection would lead to say 'x' being a complex number.

So if you sub x=a+bi into any of the equations you will see that x² gives a complex number as well, which would lead to 'y' being complex as well.

 

[femas]

Thank you for the reply.

But 'x' may be real but greater than R which makes 'y' to be pure imaginary.

So I don't know what is the case that makes 'x' real?

 

[rock.freak667]

Thank you for the reply.

But 'x' may be real but greater than R which makes 'y' to be pure imaginary.

So I don't know what is the case that makes 'x' real?

Well if you are plotting on the cartesian plane, then I don't think you would see a geometric reason, if you perhaps plot it on an Argand diagram, you might see a geometric reason.

 

[D H]

No intersection would lead to say 'x' being a complex number.

Sure it would. Let's look at

\aligned<br /> \phantom1 x^2+\phantom1 y^2 &amp;= \phantom1 1 \\<br /> 9x^2 + 4y^2 &amp;= 36<br /> \endaligned

The solutions (there are four of them) are given by x^2=32/5, y^2=-27/5. Note that in this case y is pure imaginary.