Proof that $(x-a,y-b)\subset\mathbb{R}[X,Y]/(X^2+Y^2-1)$ is maximal iff $a^2+b^2=1$ 
Let $R=\mathbb{R}[X,Y]/I, I=(X^2+Y^2-1)$ and let $x=X \mod{I}$, $y=Y\mod{I}\in R$.
a.  Proof that $(x-a,y-b)\subset\mathbb{R}[X,Y]/(X^2+Y^2-1)$, $a,b\in\mathbb{R}$ is a maximal ideal iff $a^2+b^2=1$.
b. For which $b\in\mathbb{R}$ is $(y-b)$ a maximal ideal?

My try: For part a, first suppose $a^2+b^2=1$. Then $(X^2+Y^2-1)\subset(X-a,Y-b)$, and thus $(X-a,Y-b)/I=(X-a\mod{I},Y-b\mod{I})=(x-a,y-b)$, and then by the third isomorphism theorem: $$R/(x-a,y-b)=(\mathbb{R}[X,Y]/I)/((X-a,Y-b)/I)\cong\mathbb{R}[X,Y]/(X-a,Y-b)\cong\mathbb{R},$$ which is a field, and thus $(x-a,y-b)\subset R$ is a maximal ideal.
For the other implication, suppose $a^2+b^2\neq1$, observe that $$(x-a,y-b)=(x-a,y-b,x^2+y^2-(a^2+b^2))=(x-a,y-b,1-(a^2+b^2)),$$ and because $1-(a^2+b^2)$ is invertible since $a^2+b^2\neq1$, $(x-a,y-b)=(1)$, and is thus not maximal.
I have no idea how to solve part b.
My question: is my solution to a correct? How can I solve part b? Any help is much appreciated.
 A: Your solution to part (a) is correct.  For part (b), I would suggest you use the same trick you used in part (a): to identify if an ideal is maximal, look at the quotient and rewrite the quotient using the third isomorphism theorem.  In particular, you can identify $R/(y-b)$ with $\mathbb{R}[X,Y]/(X^2+Y^2-1, Y-b)$.  Can you find a simpler way to write this quotient?
More details are hidden below.

 We can mod out $Y-b$ first, to find that $$\mathbb{R}[X,Y]/(X^2+Y^2-1, Y-b)\cong \mathbb{R}[X]/(X^2+b^2-1).$$ Now, since $\mathbb{R}[X]$ is a PID, this quotient is a field iff $X^2+b^2-1$ is irreducible.  I'll leave it to you to identify for which $b$ that is true.

A: For part (b), let $(y-b)\subset R$ be a maximal ideal, than $(Y-b)\subset \mathbb{R}[X,Y]$ is an ideal (third isomorfism theorem). Thus there is a homomorfism $\varphi$ with $\ker(\varphi)=(Y-b)$, e.g. $\varphi:f\mapsto f(0,b)$, because every ideal is a kernel of a homomorfism. We know $(X^2+Y^2-1)\subset(Y-b)$ and thus $(X^2+Y^2-1)\subset\ker(\varphi)$, this implies $\varphi(X^2+Y^2-1) = 0$, thus $b^2 = 1$.
