Finding the Points of Contact of Two Circles Using Coaxial Circle and Coordinate Geometry
Introduction
Discovering the points of contact between two circles is a classic problem in geometry. This article discusses a method to solve such problems using the properties of coaxial circles and coordinate geometry. The specific focus will be on two circles given by the equations:
Circle 1: (x^2 y^2 - 6y - 8 0) Circle 2: (x^2 y^2 - 12x - 10y - 60 0)Coaxial Circles Approach
The concept of coaxial circles is crucial for solving this problem. Coaxial circles are circles sharing a common radical axis, and they play a vital role in determining the points of contact.
Step 1: Express the general form of the coaxial circles
A general coaxial circle can be represented as:
[lambda (x^2 y^2 - 6y - 8) - (x^2 y^2 - 12x - 10y - 60) 0]Expanding and simplifying the equation:
[lambda x^2 lambda y^2 - 6lambda y - 8lambda - x^2 - y^2 12x 10y 60 0] [Rightarrow (lambda - 1)x^2 (lambda - 1)y^2 - 6lambda y 12x 10y 60 - 8lambda 0]Step 2: Completing the square
For circles to be co-axial and of zero radius, we must find the value of (lambda). The point of contact lies at the center of the coaxial circle:
[text{Center} left(frac{6lambda}{1 - lambda}, frac{10lambda 60}{1 - lambda}right)]Solving for (lambda), we get (lambda -frac{11}{10}). Substituting this value gives us the coordinates of the point of contact:
[text{Center/Point of contact} left(frac{6left(-frac{11}{10}right)}{1 - left(-frac{11}{10}right)}, frac{10left(-frac{11}{10}right) 60}{1 - left(-frac{11}{10}right)}right)] [Rightarrow left(frac{-frac{66}{10}}{frac{21}{10}}, frac{-11 60}{1 frac{11}{10}}right) left(frac{-66}{21}, frac{49}{21}right) left(-frac{3}{5}, -frac{19}{5}right)]Alternative Approach: Parametric Form
Another approach is to use parametric forms of the circles. The first circle has a center at ((0, -3)) and radius 1. Hence, its parametric form is:
[begin{cases} x cos alpha y -3 sin alpha end{cases}]The second circle has a center at ((6, 5)) and radius 11. Therefore, its parametric form is:
[begin{cases} x 6 11 cos beta y 5 11 sin beta end{cases}]Setting these equal gives:
[cos alpha 6 11 cos beta] [ -3 sin alpha 5 11 sin beta]Solving these, we obtain:
[11 cos beta cos alpha - 6 quad text{and} quad 11 sin beta sin alpha - 8]Squaring and adding the equations:
[(-3 sin alpha - 8)^2 (cos alpha - 6)^2 1]Using trigonometric identities, we can solve for (cos alpha) and (sin alpha), leading to the point of contact:
[cos alpha -frac{3}{5} quad text{and} quad sin alpha -frac{4}{5}] [x -3/5, quad y -3 - 4/5 -3.8]Algebraic Solution
We can also solve the problem algebraically by subtracting the equations of the circles:
[(x^2 y^2 - 6y - 8) - (x^2 y^2 - 12x - 10y - 60) 0] [Rightarrow 12x 16y -68 quad text{or} quad 3x 4y -17]Solving for (x) in terms of (y):
[x frac{-17 - 4y}{3}]Substitute (x) back into one of the original circle equations and solve for (y):
[y^2 left(frac{-17-4y}{3}right)^2 - 6y - 8 0] [Rightarrow 25y^2 190y 361 0] [Rightarrow y -frac{19}{5}, quad x -frac{3}{5}]Conclusion
Both methods confirm that the point of contact between the two circles is (left(-frac{3}{5}, -frac{19}{5}right)right)). This shows how the unique co-linear nature of the centers and the point of contact ensures a unique solution.
-
The Evolution and Invention of the Spanish Language: Beyond Columbus
The Evolution and Invention of the Spanish Language: Beyond Columbus The Spanish
-
Exploring the Celtic Heritage: The Traditional and Influential Lands of the Celts
Exploring the Celtic Heritage: The Traditional and Influential Lands of the Celt