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Use Newton's method for a system to write $x^2+y^2=25$ and $x^2-y=2$ in the form $J*\delta=-f$. Define the matrix $J$ and vectors delta, $f$. Dont perform iterations.
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$f$ is given by:
$$f(x,y) = \begin{bmatrix} f_1(x,y) \\ f_2(x,y) \end{bmatrix} = \begin{bmatrix} x^2+y^2-25 \\ x^2-y-2\end{bmatrix}$$
The Jacobian is given by:
$$J_{(x,y)}(f)=\begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{bmatrix} = \begin{bmatrix} 2x & 2y \\ 2x & -1 \end{bmatrix}.$$
We are asked to write out the Newton iteration in the form $J~\delta=-f$. This yields:
$$ J~\delta=-f \implies \begin{bmatrix} 2x & 2y \\ 2x & -1 \end{bmatrix} \cdot \begin{bmatrix} x^{(k+1)}-x^{(k)} \\ y^{(k+1)}-y^{(k)} \end{bmatrix} =- \begin{bmatrix} (x^{(k)})^2 + (y^{(k)})^2 -25 \\ (x^{(k)})^2 - y^{(k)} -2 \end{bmatrix}.$$
Although you are not asked to do this, you would choose an initial guess $(x^{(0)},y^{(0)})$ and iterate to some prescribed tolerance.
You could then compare the numerical solution to the exact results, which are:
$$x = \pm \sqrt{\dfrac 12 (3+\sqrt{93})}, ~~ y = \dfrac 12 (\sqrt{93}-1)$$
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