In the realm of Pure Mathematics, particularly within coordinate geometry, the concept of graph intersections is pivotal. This section aims to provide a comprehensive understanding of the relationship between graph intersection points and the solutions of equations. It explores the conditions under which a line intersects, touches, or does not meet a curve, employing both algebraic methods and graphical illustrations.

**Intersection Points and Equation Solutions**

**Intersection Points**: These are the points where two or more graphs meet. They are crucial as they represent the solutions to the system of equations depicted by the graphs.**Graphical Solution Method**: To find solutions to a system of equations, plot each equation on the same set of axes. The coordinates of the intersection points are the solutions.

**Conditions for Intersection**

**Lines and Curves**: A line can intersect a curve at no point, exactly one point (tangent), or multiple points.**Determining Intersection Conditions**: Algebraically solving the equations simultaneously helps determine the number of intersection points.

**Lines Intersecting Curves**

**Tangent Lines**: A tangent line touches a curve at exactly one point without crossing it. The slope of the tangent at any point on a curve equals the derivative of the curve's equation at that point.**Normal Lines**: A normal line is perpendicular to a tangent at the point of tangency.

**Algebraic Methods for Finding Intersections**

**Simultaneous Equations**: Intersection points are found by solving the equations simultaneously.

**Example**:

Determine the intersection points of the line $y = x + 1$ and the circle $x^2 + y^2 = 4$.

#### Solution:

**Solving for**$x$: The solutions are $x = -\frac{1}{2} + \frac{\sqrt{7}}{2}$ and $x = -\frac{1}{2} - \frac{\sqrt{7}}{2}$.**Finding**$y$-Coordinates:- For $x = -\frac{1}{2} + \frac{\sqrt{7}}{2}$, $y = \frac{1}{2} + \frac{\sqrt{7}}{2}$.
- For $x = -\frac{1}{2} - \frac{\sqrt{7}}{2}$, $y = \frac{1}{2} - \frac{\sqrt{7}}{2}$.

**Intersection Points**:- Point 1: $(-1/2 + \sqrt{7}/2, 1/2 + \sqrt{7}/2)$
- Point 2: $(-1/2 - \sqrt{7}/2, 1/2 - \sqrt{7}/2)$

These points are where the line $y = x + 1$ and the circle $x^2 + y^2 = 4$intersect.

**Practical Problems**

### Problem 1:

Find the intersection points of the line $y = 2x - 3$ and the circle $x^2 + y^2 = 25$.

#### Solution:

**1. Combine Equations**:

Substituting $y$ from the line equation into the circle equation results in $5x^2 - 12x - 16 = 0$.

**2. Solve for **$x$:

The quadratic equation is solved to find $x$ values:

- $x = \frac{6}{5} - \frac{2\sqrt{29}}{5}$ and $x = \frac{6}{5} + \frac{2\sqrt{29}}{5}$.

**3. Find **$y$ Values:

Corresponding $y$ values are calculated as:

- For $x = \frac{6}{5} - \frac{2\sqrt{29}}{5}$, $y = -\frac{4\sqrt{29}}{5} - \frac{3}{5}$.
- For $x = \frac{6}{5} + \frac{2\sqrt{29}}{5}$, $y = \frac{4\sqrt{29}}{5} - \frac{3}{5}$.

**4. Intersection Points**:

**Point 1:**$\left(\frac{6}{5} - \frac{2\sqrt{29}}{5}, -\frac{4\sqrt{29}}{5} - \frac{3}{5}\right)$**Point 2:**$\left(\frac{6}{5} + \frac{2\sqrt{29}}{5}, \frac{4\sqrt{29}}{5} - \frac{3}{5}\right)$

These points are where the given line and circle intersect.

**Problem 2**:

Check if $y = -3x + 4$ is tangent to $x^2 + y^2 = 10$.

**Solution**:

**1. Line Equation**: $y = -3x + 4$

**2. Circle Equation**: $x^2 + y^2 = 10$

**3. Substitute **$y$** into Circle Equation**:

$x^2 + (-3x + 4)^2 = 10$

$x^2 + 9x^2 - 24x + 16 = 10$

$10x^2 - 24x + 6 = 0$

**4. Solve Quadratic Equation for **$x$:

$a = 10, b = -24, c = 6$

$x = \frac{-(-24) \pm \sqrt{(-24)^2 - 4(10)(6)}}{2(10)}$

$x = \frac{24 \pm \sqrt{576 - 240}}{20}$

$x = \frac{24 \pm \sqrt{336}}{20}$

$x = \frac{24 \pm \sqrt{84}}{10}$

**5. Check for Tangency**: A line is tangent to a circle if the quadratic equation has exactly one solution, which occurs when the discriminant $(b^2 - 4ac)$ is zero.

Discriminant $= (-24)^2 - 4(10)(6) = 576 - 240 = 336$

Since the discriminant is not zero, the equation has two distinct solutions, indicating two intersection points between the line and the circle.

**Conclusion**: The line $y = -3x + 4$ is not tangent to the circle $x^2 + y^2 = 10$ as there are two intersection points.

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.