Find the radical of${\langle x^2-y^2,x^4-x^2y^5\rangle}\subset \mathbb{C}[x,y]$

Question

Find the radical of $J:=\langle x^2-y^2,x^4-x^2y^5\rangle\subset \mathbb{C}[x,y]$.

I would like to verify my answer please;

Edit after comments:

Excluding $(i,-1)$ from $V(J)$ and $$\sqrt J=\langle (x-a)(y-b):(a,b)\in V(J)\rangle$$

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One can calculate and observe $$V(J)=\{(0,0),(1,1,),(-1,1),(i,-1),(-i,-1)\}$$ According to the Nullstellensatz $$\sqrt J=I(V(J))\\=\langle (x-a)^i(y-b)^j:i,j\in\mathbb{N},(a,b)\in V(J)\rangle$$

Answer 1

If $(a,b)\in V(J)$ then $a^2=b^2$ and hence also $$0=a^4-a^2b^5=b^4-b^7=b^4(1-b^3),$$ so either $b=0$ or $b$ is a third root of unity, and hence with $\zeta\in\Bbb{C}$ a primitive third root of unity $$V(J)\subset\{(0,0)\}\cup\{(\pm\zeta^k,\zeta^k)\},$$ and it is not hard to check that equality holds. Now the Nullstellensatz gives you $\sqrt{J}$.