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Intersection of a Line and a Circle Simplified Revision Notes

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Intersection of a Line and a Circle

What is the Intersection of a Line and a Circle?

The intersection of a line and a circle can result in:

  1. No Intersection: The line does not touch the circle.
  2. One Point of Intersection: The line is tangent to the circle.
  3. Two Points of Intersection: The line crosses through the circle.

Determining the Intersection Points

To find the points of intersection between a line and a circle:

Substitute the Equation of the Line into the Circle:

  • Equation of a circle: (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2
  • Equation of a line: y=mx+cy = mx + c

Solve the Resulting Quadratic Equation:

  • Substitute y=mx+cy = mx + c into (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2
  • This gives a quadratic equation in xx

Analyse the Discriminant (Δ\Delta):

  • Δ=b24ac\Delta = b^2 - 4ac, where a,b,ca, b, c are coefficients from the quadratic equation.
  • Δ>0\Delta > 0: Two points of intersection.
  • Δ=0\Delta = 0: One point of intersection (tangent).
  • Δ<0\Delta < 0: No intersection.

Worked Examples

infoNote

Example 1: Find Points of Intersection

Problem: Find the points of intersection between the circle x2+y2=25x^2 + y^2 = 25 and the line y=2x1y = 2x - 1.


Solution:

Step 1: Substitute y=2x1y = 2x - 1 into x2+y2=25x^2 + y^2 = 25:

x2+(2x1)2=25x^2 + (2x - 1)^2 = 25

Step 2: Expand and simplify:

x2+4x24x+1=25x^2 + 4x^2 - 4x + 1 = 255x24x24=05x^2 - 4x - 24 = 0

Step 3: Solve the quadratic equation using the quadratic formula:

x=(4)±(4)24(5)(24)2(5)=4±16+48010x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(5)(-24)}}{2(5)} = \frac{4 \pm \sqrt{16 + 480}}{10}x=4±49610=4±43110=2±2315x = \frac{4 \pm \sqrt{496}}{10} = \frac{4 \pm 4\sqrt{31}}{10} = \frac{2 \pm 2\sqrt{31}}{5}

Step 4: Substitute xx values back into y=2x1y = 2x - 1 to find yy coordinates:

y1=2(2+2315)1y_1 = 2\left(\frac{2 + 2\sqrt{31}}{5}\right) - 1y2=2(22315)1y_2 = 2\left(\frac{2 - 2\sqrt{31}}{5}\right) - 1

Answer: Points of intersection are:

(2+2315,4+43151),(22315,443151)\left(\frac{2 + 2\sqrt{31}}{5}, \frac{4 + 4\sqrt{31}}{5} - 1\right), \quad \left(\frac{2 - 2\sqrt{31}}{5}, \frac{4 - 4\sqrt{31}}{5} - 1\right)

infoNote

Example 2: Determine Tangency

Problem: Show that the line y=12x+3y = -\frac{1}{2}x + 3 is tangent to the circle x2+y2=10x^2 + y^2 = 10.


Solution:

Step 1: Substitute y=12x+3y = -\frac{1}{2}x + 3 into x2+y2=10x^2 + y^2 = 10:

x2+(12x+3)2=10x^2 + \left(-\frac{1}{2}x + 3\right)^2 = 10

Step 2: Expand and simplify:

x2+14x23x+9=10x^2 + \frac{1}{4}x^2 - 3x + 9 = 1054x23x1=0\frac{5}{4}x^2 - 3x - 1 = 0

Step 3: Calculate the discriminant (Δ\Delta):

Δ=(3)24(54)(1)=9+5=14\Delta = (-3)^2 - 4\left(\frac{5}{4}\right)(-1) = 9 + 5 = 14

Answer: The discriminant is 00, confirming the line is tangent.


Summary

  • Steps for Finding Intersection:
    1. Substitute the line's equation into the circle's equation.
    2. Solve the resulting quadratic equation.
    3. Use the discriminant (Δ\Delta) to classify the intersection.
  • Key Cases:
    • Δ>0\Delta > 0: Two points of intersection.
    • Δ=0\Delta = 0: Tangency (one point).
    • Δ<0\Delta < 0: No intersection.
  • Practice solving such problems to understand tangency and intersections with circles.
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