By solving equations, find the exact points at which the circle x^2+ y^2= 4^2 intersects the straight line y = 2x.

To find the exact points at which the circle \( x^2 + y^2 = 4^2 \) intersects the straight line \( y = 2x \), we can solve the system of equations step by step.

### Step 1: Write down the equations.

The equation of the circle is:

\[
x^2 + y^2 = 4^2
\]

This simplifies to:

\[
x^2 + y^2 = 16
\]

The equation of the line is:

\[
y = 2x
\]

### Step 2: Substitute the equation of the line into the equation of the circle.

Substitute \( y = 2x \) into the circle's equation:

\[
x^2 + (2x)^2 = 16
\]

### Step 3: Simplify and solve the quadratic equation.

Simplify the equation:

\[
x^2 + 4x^2 = 16
\]

\[
5x^2 = 16
\]

\[
x^2 = \frac{16}{5}
\]

\[
x = \pm \sqrt{\frac{16}{5}} = \pm \frac{4}{\sqrt{5}} = \pm \frac{4\sqrt{5}}{5}
\]

### Step 4: Find the corresponding \( y \) values.

Since \( y = 2x \), substitute the \( x \) values back into the equation for \( y \):

For \( x = \frac{4\sqrt{5}}{5} \):

\[
y = 2 \times \frac{4\sqrt{5}}{5} = \frac{8\sqrt{5}}{5}
\]

For \( x = -\frac{4\sqrt{5}}{5} \):

\[
y = 2 \times \left(-\frac{4\sqrt{5}}{5}\right) = -\frac{8\sqrt{5}}{5}
\]

### Step 5: Write the intersection points.

The exact points at which the circle intersects the line are:

\[
\left(\frac{4\sqrt{5}}{5}, \frac{8\sqrt{5}}{5}\right) \quad \text{and} \quad \left(-\frac{4\sqrt{5}}{5}, -\frac{8\sqrt{5}}{5}\right)
\]

Would you like further details on this solution, or do you have any questions?

Here are some related questions:
1. How do you find the slope of a line given its equation?
2. What is the significance of the radius in the equation of a circle?
3. How do you determine if a point lies on a given circle?
4. How can you solve for intersection points between a circle

Explore how to find the exact points of intersection between the circle x^2 + y^2 = 16 and the line y = 2x. Detailed step-by-step solution includes substitution, simplification, and solving quadratic equations. Suitable for high school students and advanced learners.

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