To solve the given system of equations:

1. $x^2 + y^2 = 106$
2. $x - y = 4$

We will determine the points of intersection by solving this system step-by-step.

Step 1: Solve the linear equation for one variable

From the second equation: $x = y + 4$

Step 2: Substitute into the first equation

Substituting $x = y + 4$ into $x^2 + y^2 = 106$: $(y + 4)^2 + y^2 = 106$

Expand $(y + 4)^2$: $y^2 + 8y + 16 + y^2 = 106$

Combine like terms: $2y^2 + 8y + 16 = 106$

Simplify: $2y^2 + 8y - 90 = 0$

Divide through by 2 to simplify: $y^2 + 4y - 45 = 0$

Step 3: Solve the quadratic equation

Factorize $y^2 + 4y - 45 = 0$: $(y + 9)(y - 5) = 0$

So: $y = -9 \quad \text{or} \quad y = 5$

Step 4: Find corresponding $x$-values

Using $x = y + 4$:

• If $y = -9$, then $x = -9 + 4 = -5$.
• If $y = 5$, then $x = 5 + 4 = 9$.

Final Answer:

The points of intersection are: $(x, y) = (-5, -9) \quad \text{and} \quad (9, 5)$

Would you like a graph of this system or further details on any step?

Here are five related questions to explore further:

1. What is the geometric interpretation of the system of equations?
2. How do we confirm these points lie on the given circle?
3. What other techniques can solve nonlinear systems of equations?
4. How would the solution change if $x - y = 4$ was replaced with $x + y = 4$?
5. Can the discriminant of the quadratic equation confirm real solutions exist?

Tip: Always simplify equations as much as possible before solving to make calculations easier!