Introduction to Circle and Straight Line Intersection

Determining the points of intersection between a circle and a straight line involves solving a system of equations. This tutorial will walk you through the process, using the given equation of a circle and a straight line to find their intersection points. This is a fundamental concept in both algebra and calculus, with applications in various fields such as engineering and physics.

Solving the Circle Equation

The given circle equation is x2 y2 - 4x - 6y 4 0. To better understand this equation, we first need to rewrite it in the standard form of a circle: ((x - h)^2 (y - k)^2 r^2), where ((h, k)) is the center of the circle and (r) is its radius.

Step 1: Grouping the Variables

We group the variables as follows:

[(x^2 - 4x) (y^2 - 6y) -4]

Step 2: Completing the Squares

To complete the squares, we add and subtract the necessary constants:

[(x^2 - 4x 4) (y^2 - 6y 9) -4 4 9]

This results in:

[(x - 2)^2 (y - 3)^2 9]

This shows that the circle has a center at ((2, 3)) and a radius of (3).

Intersection with the Straight Line

Next, we are given the straight line equation as (y x). We need to find the points of intersection between the circle and this line.

Step 1: Substituting (y x)

Substitute (y x) into the circle's equation:

[x^2 x^2 - 4x - 6x 4 0]

This simplifies to:

[2x^2 - 1 4 0]

Step 2: Solving the Quadratic Equation

To solve the quadratic equation (2x^2 - 1 4 0), we can use the quadratic formula (x frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a 2), (b -10), and (c 4).

Calculate the discriminant:

[b^2 - 4ac (-10)^2 - 4 cdot 2 cdot 4 100 - 32 68]

Therefore, the solutions for (x) are:

[x frac{10 pm sqrt{68}}{4}]

Since (sqrt{68} sqrt{4 cdot 17} 2sqrt{17}), the solutions can be simplified to:

[x frac{10 pm 2sqrt{17}}{4} frac{5 pm sqrt{17}}{2}]

This gives us two x-values:

[x_1 frac{5 sqrt{17}}{2}, quad x_2 frac{5 - sqrt{17}}{2}]

Step 3: Calculating the Corresponding y-values

Since (y x), the corresponding y-values are the same as the x-values. Therefore, the points of intersection are:

[(x, y) left(frac{5 sqrt{17}}{2}, frac{5 sqrt{17}}{2}right) quad text{and} quad left(frac{5 - sqrt{17}}{2}, frac{5 - sqrt{17}}{2}right)]

Conclusion

In summary, we have found the points of intersection between the circle (x^2 y^2 - 4x - 6y 4 0) and the straight line (y x). These points are:

[left(frac{5 sqrt{17}}{2}, frac{5 sqrt{17}}{2}right) quad text{and} quad left(frac{5 - sqrt{17}}{2}, frac{5 - sqrt{17}}{2}right)]

This process involved algebraic substitution and solving a quadratic equation, providing a clear example of how to handle geometric problems analytically.