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Define bilinear transformation and prove that in a general a bilinear transformation maps into a circle.
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Definition:

The bilinear transformation is a mathematical mapping of variables. In digital filtering, it is a standard method of mapping the s or analog plane into the z or digital plane. It transforms analog filters, designed using classical filter design techniques, into their discrete equivalents.

Proof: Let the bilinear transformation be

$w = \frac{az + b}{cz + d} \\ \therefore w = \frac{a}{c} + \frac{b - \frac{da}{c}}{cz + d}$ (writing in standard form)

Simplifying we get

$\frac{a}{c} + \frac{bc - ba}{c} * \frac{1}{cz + d} \\ \frac{a}{c} + \frac{bc - da}{c^2} * \frac{1}{\bigg(z + \frac{d}{c}\bigg)}$

As above, by actual division, we can write it as

$w = \frac{a}{c} + \frac{bc - ba}{c^2} * \frac{1}{\bigg(z + \frac{d}{c}\bigg)}$

Now, consider the transformation $w_{1} = z + \frac{d}{c}$ which is a translation.

Since in the transformation w = z + k size and shape is preserved, circles in z-plane will be transformed into circles in $w_{1}$ plane. Now, consider the transformation $w_{2} = \frac{1}{w_{1}}$, which is inversion and reflection.

Since in the transformation $w = \frac{1}{z}$ circles are mapped onto circles, circles in $w_{1}$ plane will be mapped onto circles in $w_{2}$ plane.

Now, consider the transformation $w_{3} = \frac{bc - ad}{c^2} * w_{2}$ which is a rotation and magnification.

Since in the transformation w = kz figures are only rotated and magnified, circles in $w_{2}$ plane will be transformed into circles in $w_{3}$ plane.

Now, consider the transformation $w = \frac{a}{c} + w_{3}$, which again is a translation.

Hence, circles in $w_{3}$ plane will be transformed into circles in the w - plane.

Hence, circles in Z - plane will be transformed into circles in w-plane.

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