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CIE A-Level Maths Study Notes

1.3.6 Intersections of Graphs

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.
intersections of graphs: circle and lines

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+1y = x + 1 and the circle x2+y2=4x^2 + y^2 = 4.

Solution:

  • Solving for xx: The solutions are x=12+72x = -\frac{1}{2} + \frac{\sqrt{7}}{2} and x=1272x = -\frac{1}{2} - \frac{\sqrt{7}}{2}.
  • Finding yy-Coordinates:
    • For x=12+72x = -\frac{1}{2} + \frac{\sqrt{7}}{2}, y=12+72y = \frac{1}{2} + \frac{\sqrt{7}}{2}.
    • For x=1272x = -\frac{1}{2} - \frac{\sqrt{7}}{2}, y=1272y = \frac{1}{2} - \frac{\sqrt{7}}{2}.
  • Intersection Points:
    • Point 1: (1/2+7/2,1/2+7/2)(-1/2 + \sqrt{7}/2, 1/2 + \sqrt{7}/2)
    • Point 2: (1/27/2,1/27/2)(-1/2 - \sqrt{7}/2, 1/2 - \sqrt{7}/2)

These points are where the line y=x+1y = x + 1 and the circle x2+y2=4x^2 + y^2 = 4intersect.

intersection of line and circle illustration

Practical Problems

Problem 1:

Find the intersection points of the line y=2x3y = 2x - 3 and the circle x2+y2=25x^2 + y^2 = 25.

Solution:

1. Combine Equations:

Substituting yy from the line equation into the circle equation results in 5x212x16=05x^2 - 12x - 16 = 0.

2. Solve for xx:

The quadratic equation is solved to find xx values:

  • x=652295x = \frac{6}{5} - \frac{2\sqrt{29}}{5} and x=65+2295x = \frac{6}{5} + \frac{2\sqrt{29}}{5}.

3. Find yy Values:

Corresponding yy values are calculated as:

  • For x=652295x = \frac{6}{5} - \frac{2\sqrt{29}}{5}, y=429535y = -\frac{4\sqrt{29}}{5} - \frac{3}{5}.
  • For x=65+2295x = \frac{6}{5} + \frac{2\sqrt{29}}{5}, y=429535y = \frac{4\sqrt{29}}{5} - \frac{3}{5}.

4. Intersection Points:

  • Point 1: (652295,429535)\left(\frac{6}{5} - \frac{2\sqrt{29}}{5}, -\frac{4\sqrt{29}}{5} - \frac{3}{5}\right)
  • Point 2: (65+2295,429535)\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.

intersection of line and circle

Problem 2:

Check if y=3x+4y = -3x + 4 is tangent to x2+y2=10x^2 + y^2 = 10.

Solution:

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

2. Circle Equation: x2+y2=10x^2 + y^2 = 10

3. Substitute yy into Circle Equation:

x2+(3x+4)2=10x^2 + (-3x + 4)^2 = 10

x2+9x224x+16=10x^2 + 9x^2 - 24x + 16 = 10

10x224x+6=010x^2 - 24x + 6 = 0

4. Solve Quadratic Equation for xx:

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

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

x=24±57624020x = \frac{24 \pm \sqrt{576 - 240}}{20}

x=24±33620x = \frac{24 \pm \sqrt{336}}{20}

x=24±8410x = \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 (b24ac)(b^2 - 4ac) is zero.

Discriminant =(24)24(10)(6)=576240=336= (-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+4y = -3x + 4 is not tangent to the circle x2+y2=10x^2 + y^2 = 10 as there are two intersection points.

tangency of line and circle
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

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.

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