## Calculus 8th Edition

The equation of the tangent line is $$y= x+\frac{1}{2}$$
Given $$x^{2}+y^{2}=\left(2 x^{2}+2 y^{2}-x\right)^{2}, \ \ \ \ \ (0,\frac{1}{2})$$ Since $x^{2}+y^{2}=\left(2 x^{2}+2 y^{2}-x\right)^{2}$ by differentiate both sides by $x$, we get \begin{align} \Rightarrow &2 x+2 y y^{\prime}=2\left(2 x^{2}+2 y^{2}-x\right)\left(4 x+4 y y^{\prime}-1\right)\\ \end{align} at $(x,y)=(0,\frac{1}{2})$, we get $0+y^{\prime}=2\left(\frac{1}{2}\right)\left(2 y^{\prime}-1\right) \Rightarrow y^{\prime}=2 y^{\prime}-1$ so, $$y^{\prime}=1$$ henc the equation of the tangent line is $y-\frac{1}{2}=1(x-0)$ i.e $y= x+\frac{1}{2}$