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Finding the Points of Contact of Two Circles Using Coaxial Circle and Coordinate Geometry

February 28, 2025Culture3986
Introduction Discovering the points of contact between two circles is

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.