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.