$$y=\frac{3}{2}x-2$$
$$y^2x+a=x^2+y^2$$
This is what I have so far. Tangent and curve are intersecting in $(0,-2)$ so
$$4*0+a=0+4 \Rightarrow a=4$$
I have to check $y'(0)=\frac{3}{2}$. I don't know how to derive curve because i get $y$ and $y'$. And if i don't get $y'(0)=\frac{3}{2}$, then $a$ doesn't exist?
On
The $x$-coordinates of the intersections of the line and curve for a given $a$ can be found by substituting the line's equation into the curve: $$y^2x+a=x^2+y^2$$ $$y^2(x-1)-x^2+a=0$$ $$\left(\frac32x-2\right)^2(x-1)-x^2+a=0$$ $$\frac94x^3-\frac{37}4x^2+10x-(4-a)=0$$ For the intersection to be tangential, the discriminant of this equation must be zero, which will create a multiple root. The discriminant with respect to $x$ is $$\Delta=-\frac1{16}(2187a^2-8209a+4944)$$ and this is zero when $$a=3\text{ or }a=\frac{1648}{2187}$$ At these values of $a$, then, the given line is tangent to the curve.
Hint: Differentiating the second equation with respect to $x$, you get $$y^2+2xyy'=2x+2yy'.$$ But the slope of $y=3x/2-2$ is $3/2$, so the value of $y'$ at the point of intersection should be $y'=3/2$ as well. Plugging this back in gives $$y^2+3xy=2x+3y,$$ and you want to find the intersection of this with $y=3x/2-2$. The rest is algebra.
For geometric clarity, here's Mathematica plots of the three curves discussed for the cases indicated by Parcly Taxel's answer:
The blue curve is the line $y=3x/2-2$, the orange curve is the intersecting curve (with $a=3$ in the first plot and $a=\frac{1684}{2187}$ in the second), and the green curve is the equation $y^2+3xy=2x+3y$. The point is that, for these values of $a$, all three curves intersect at a common point, and the blue and orange curves are tangent at this point.