MHB How can I find the center and radius of a circle from two different equations?

[aruwin]

Hi,
how do I find the center and radius from these equations? The 2 equations represent 2 different circles, by the way.

 

[chisigma]

Hi,
how do I find the center and radius from these equations? The 2 equations represent 2 different circles, by the way.

The only visible function and the exponential function...

$\displaystyle e^{z} = \sum_{n=0}^{\infty} \frac{z^{n}}{n!}\ (1)$

... which has centre in z=0 and converges for any value of z...

Kind regards

$\chi$ $\sigma$

 

[aruwin]

The only visible function and the exponential function...

$\displaystyle e^{z} = \sum_{n=0}^{\infty} \frac{z^{n}}{n!}\ (1)$

... which has centre in z=0 and converges for any value of z...

Kind regards

$\chi$ $\sigma$

I forgot to mention that P+Q is a complex number in the form Z= X+iY because P is the real power while Q is the reactive power.

 

[DavidCampen]

I will work the 1st equation. I will eliminate subscripts to make the notation simpler and also let \(\alpha = \theta - \frac{\pi}{2} \) and \(K = \frac{E^2}{X}\)

So the 1st equation becomes:
\(P + Q = K(\,(0.9)^2j + 0.9e^{j\alpha}\,)\)

we know that \(e^{j\alpha} = \cos(\alpha) + j \sin(\alpha)\)

so the 1st equation becomes:
\(P + Q = K[\,(0.9)^2j + 0.9(\,\cos(\alpha) + j \sin(\alpha)\,)]\)

or
\(P + Q = 0.9K(\,\cos(\alpha) + j \sin(\alpha)\,) + (0.9)^2Kj \)

so \(P + Q\) as a function of \(\alpha\) is a circle of radius \(0.9K\) and center (\(0.9)^2Kj \)