Answer
The solutions of the stated algebraic equations are $\left( 3,4 \right)$ and$\left( 3,-4 \right)$.
Work Step by Step
The equations are as given below:
${{x}^{2}}+{{y}^{2}}=25$ (I)
${{\left( x-8 \right)}^{2}}+{{y}^{2}}=41$ (II)
Now, demonstrate the steps as given below:
Step 1:
When necessary, multiply either of the equations or both equations by a significant number so that the sum of the coefficients of ${{x}^{2}}$ or the sum of the coefficients of ${{y}^{2}}$ is zero.
Thus, multiply the equation (II) by the value $-1$ so that the sum of the coefficients of ${{y}^{2}}$ becomes zero.
$\begin{align}
& -{{x}^{2}}-{{y}^{2}}+\left( {{\left( x-8 \right)}^{2}}+{{y}^{2}} \right)=-25+41 \\
& {{\left( x-8 \right)}^{2}}-{{x}^{2}}=16 \\
& {{x}^{2}}+64-16x-{{x}^{2}}=16
\end{align}$
$\begin{align}
& 48=16x \\
& 3=x \\
\end{align}$
Therefore, $x=3$ is a solution of both equations.
Step 2:
Substitute the value of $x$ in equation (I) to find out the value of y as given below:
For $x=3$,
$\begin{align}
& {{x}^{2}}+{{y}^{2}}=25 \\
& {{3}^{2}}+{{y}^{2}}=25 \\
& 9+{{y}^{2}}=25 \\
& {{y}^{2}}=16
\end{align}$
$\begin{align}
& {{y}^{2}}=16 \\
& {{y}^{2}}-16=0 \\
& (y-4)(y+4)=0
\end{align}$
Putting the factors as zero to get:
$\begin{align}
& y-4=0 \\
& y=4
\end{align}$
Or
$\begin{align}
& y+4=0 \\
& y=-4
\end{align}$
Therefore, $y=\pm 4$ are solutions of this equation.
Step 3:
And verify the values of $x$ and $y$ in both of the equations: start by taking the pair$\left( 3,4 \right)$ and put $x=+\sqrt{5}\text{or}-\sqrt{5}$ and $y=4$.
$\begin{align}
& {{x}^{2}}+{{y}^{2}}=25 \\
& {{3}^{2}}+{{4}^{2}}=25 \\
& 9+16=25 \\
& 25=25
\end{align}$
So, the values satisfy the equation.
And,
$\begin{align}
& {{\left( x-8 \right)}^{2}}+{{y}^{2}}=41 \\
& {{\left( 3-8 \right)}^{2}}+{{4}^{2}}=41 \\
& 25+16=41 \\
& 41=41
\end{align}$
So, the values satisfy the equation.
Step 4:
Now, begin by checking$\left( 3,-4 \right)$, and replace $x=3$ and $y=4$ .
$\begin{align}
& {{x}^{2}}+{{y}^{2}}=25 \\
& {{3}^{2}}+{{\left( -4 \right)}^{2}}=25 \\
& 9+16=25 \\
& 25=25
\end{align}$
Therefore, the values satisfy the equation.
And,
$\begin{align}
& {{\left( x-8 \right)}^{2}}+{{y}^{2}}=41 \\
& {{\left( 3-8 \right)}^{2}}+{{\left( -4 \right)}^{2}}=41 \\
& 25+16=41 \\
& 41=41
\end{align}$
Therefore, the values satisfy the equation.
Hence, the points $\left( 3,4 \right)$ and $\left( 3,-4 \right)$ are a solution of the system.
