The given system of equations consists of:

1. $x^2 + (y - 4)^2 = 90$
2. $x + 2y = -7$

This represents a circle in the first equation and a straight line in the second equation. The goal is to find the intersection points of these two equations.

Step 1: Solve the second equation for $x$:

From $x + 2y = -7$: $x = -7 - 2y$

Step 2: Substitute $x = -7 - 2y$ into the first equation:

Substituting $x = -7 - 2y$ into $x^2 + (y - 4)^2 = 90$: $(-7 - 2y)^2 + (y - 4)^2 = 90$

Step 3: Expand and simplify:

Expand $(-7 - 2y)^2$: $(-7 - 2y)^2 = 49 + 28y + 4y^2$

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

Combine: $49 + 28y + 4y^2 + y^2 - 8y + 16 = 90$

Simplify: $5y^2 + 20y + 65 = 90$

Simplify further: $5y^2 + 20y - 25 = 0$

Step 4: Divide through by 5:

$y^2 + 4y - 5 = 0$

Step 5: Solve the quadratic equation:

Factorize $y^2 + 4y - 5 = 0$: $(y + 5)(y - 1) = 0$

Thus: $y = -5 \quad \text{or} \quad y = 1$

Step 6: Solve for $x$ for each $y$:

For $y = -5$: $x = -7 - 2(-5) = -7 + 10 = 3$

For $y = 1$: $x = -7 - 2(1) = -7 - 2 = -9$

Final Answer:

The intersection points are: $(3, -5) \quad \text{and} \quad (-9, 1)$

Would you like further details or visual representation of these solutions?

Here are 5 related questions:

1. How do we interpret the intersection points geometrically?
2. What happens if the circle and line do not intersect?
3. Can we verify the solutions by substituting them back into the equations?
4. How can we solve the system using a graphing method?
5. How do we generalize this approach for different conic sections and lines?

Tip: Always check your solutions by substituting them back into the original equations to confirm they satisfy both!