Your real problem is that this curve is NOT a circle! The tangent line to any smooth curve, at a given point, is a line that goes through that point and has slope equal to the **derivative** at that point. The derivative of y, given by \(\displaystyle x^2y^2- 2x^2+ y^2= 0\), is given by \(\displaystyle 2xy^2- 2x^2y \frac{dy}{dx}- 4x+2y\frac{dy}{dx}= 0\). At the point (1, 1), setting x= y= 1, \(\displaystyle 2- 2\frac{dy}{dx}- 4+ 2\frac{dy}{dx}= 0\) so \(\displaystyle 2- 4= 0\). The \(\displaystyle \frac{dy}{dx}\) completely cancels out leaving a untrue equation! That means that the derivative does not exist there which, in turn, means the tangent line is **vertical**. What is the equation of a vertical line that passes through (1, 1)?

(You seem to be saying that you thought that only **circles** can have "tangent lines". If so, you have missed a lot of material that is preliminary to problems like this. You need to talk to your teacher about this.)