The problem in the image asks:

"Given a circle with the equation $x^2 + y^2 = 25$, find the coordinates of the points where the circle intersects the line $y = 2x + 1$."

Solution:

We have two equations to solve simultaneously:

1. The equation of the circle: $x^2 + y^2 = 25$
2. The equation of the line: $y = 2x + 1$

Step 1: Substitution

Since $y = 2x + 1$, substitute this into the circle equation.

$x^2 + (2x + 1)^2 = 25$

Step 2: Expand the equation

Expand the square on the left-hand side:

$x^2 + (4x^2 + 4x + 1) = 25$

This simplifies to:

$x^2 + 4x^2 + 4x + 1 = 25$

Combine like terms:

$5x^2 + 4x + 1 = 25$

Step 3: Rearrange the equation

Move everything to one side to set the equation equal to zero:

$5x^2 + 4x + 1 - 25 = 0$

$5x^2 + 4x - 24 = 0$

Step 4: Solve the quadratic equation

We can solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 5$, $b = 4$, and $c = -24$.

$x = \frac{-4 \pm \sqrt{4^2 - 4(5)(-24)}}{2(5)}$

$x = \frac{-4 \pm \sqrt{16 + 480}}{10}$

$x = \frac{-4 \pm \sqrt{496}}{10}$

$x = \frac{-4 \pm 22.27}{10}$

Step 5: Calculate the roots

Now, solve for both possible values of $x$:

1. $x_1 = \frac{-4 + 22.27}{10} = \frac{18.27}{10} = 1.827$
2. $x_2 = \frac{-4 - 22.27}{10} = \frac{-26.27}{10} = -2.627$

Step 6: Find the corresponding $y$ values

Now that we have the $x$-values, use the line equation $y = 2x + 1$ to find the corresponding $y$-values:

For $x_1 = 1.827$:

$y_1 = 2(1.827) + 1 = 3.654 + 1 = 4.654$

For $x_2 = -2.627$:

$y_2 = 2(-2.627) + 1 = -5.254 + 1 = -4.254$

Step 7: Final answer

The points of intersection are approximately: $(1.827, 4.654) \quad \text{and} \quad (-2.627, -4.254)$

Would you like any more details or clarifications on this?

Here are 5 follow-up questions for further exploration:

1. How do you solve a quadratic equation using factoring instead of the quadratic formula?
2. What geometric properties of circles make the solution of these kinds of problems possible?
3. How do we know that the line will intersect the circle at exactly two points?
4. Could this system have had no solutions? Under what conditions would that happen?
5. How would the solution change if the circle had a different radius?

Tip: Always simplify the equations as much as possible before applying formulas to reduce errors and computation complexity.