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We are given a polar equation $$r^2 = 16 \cos \theta$$. Recognize that this is a well-known polar equation representing a circle. We can deduce that it's a circle since the equation is of the form $$r^2 = a^2 \cos \theta$$, where a is a constant value, in our case, 16.

Now, let's convert the polar equation into a cartesian equation. Recall the cartesian conversion equations: $$x = r \cos \theta$$ $$y = r \sin \theta$$ First, let's find $$r^2$$, from the given polar equation we have $$r^2 = 16\cos \theta$$. But we know that $$r^2 = x^2 + y^2$$ (from the conversion equations), so: $$x^2 + y^2 = 16 \cos \theta$$ Now, we need to replace $$\cos \theta$$ with its cartesian equation representation, we know that $$\cos \theta = \frac{x}{r}$$. Since we want to eliminate $$r$$, we will use the equation $$r^2 = x^2 + y^2$$, and therefore $$r = \sqrt{x^2 + y^2}$$, so: $$\cos \theta = \frac{x}{\sqrt{x^2 + y^2}}$$ Now substitute this into the equation: $$x^2 + y^2 = 16 \left(\frac{x}{\sqrt{x^2 + y^2}}\right)$$

To graph the cartesian equation, use a graphing tool like Desmos. Insert the cartesian equation into the plotting tool: $$x^2 + y^2 = 16\left(\frac{x}{\sqrt{x^2 + y^2}}\right)$$ The graphing tool will display a circle after evaluating this equation, which confirms our assessment of the shape of the given polar equation. To check the accuracy of the graph, we can also graph the original polar equation ($$r^2 = 16 \cos \theta$$) using the graphing tool's polar graphing capabilities. The polar and cartesian graphs should overlap, which would confirm that our work is accurate.