Problem: For which value $a$ line $y=\frac{3}{2}x-2$ is a tangent line to this curve: $y^2x+a=x^2+y^2$
My idea:
Use implicit differentiation to find fist derivate of function.
$2yy'x+y^2=2x+2yy'$ and I get for $y'$ this: $y'=\frac{2x-y^2}{2yx-2y}$.
We know that if $t...y=ax+b$ is a tangent line of function $f(x)$ that $f'(x)=a$ at point intersection of line and a function.
So $\frac{2x-y^2}{2yx-2y}=\frac{3}{2}$.
After solving this equation I get $x=\frac{3y-y^2}{3y-2}$.
$a=x^2+y^2+y^2x$ - in this equation we can right side write in terms of y and get what y in terms of a. But that didn't help me to solve the problem.
Let $(s,t)$ be the tangent point. Then, we have $$t=\frac 32s-2\tag1$$ $$t^2s+a=s^2+t^2\tag2$$ Also, since we have $2yy'x+y^2=2x+2yy'$, $$2t\times\frac 32\times s+t^2=2s+2t\times \frac 32\iff 3st+t^2=2s+3t\tag3$$
Eliminating $t$ from $(1)(3)$ gives $$3s\left(\frac 32s-2\right)+\left(\frac 32s-2\right)^2=2s+3\left(\frac 32s-2\right),$$ i.e. $$\frac 14(s-2)(27s-20)=0\quad\Rightarrow \quad s=2,\frac{20}{27}$$ Hence, from $(1)(2)$, we have $$(s,t,\color{red}{a})=\left(2,1,\color{red}{3}\right),\quad\left(\frac{20}{27},-\frac 89,\color{red}{\frac{1648}{2187}}\right)$$