Introduction

When tackling problems related to lines and circles in geometry, one often encounters scenarios where a line is specified by its slope and its distance from the origin. This article will guide you through the process of finding the equation of such a line, using a detailed example. We will employ both algebraic and geometric methods to provide a comprehensive understanding.

Algebraic Method

Given the slope of a line is 5 and it is 9 units away from the origin, we need to find the equation of this line.

1. Understanding the Distance: Points that are 9 units away from the origin lie on a circle centered at the origin with a radius of 9. The equation of this circle is (x^2 y^2 81).

2. Equation of the Line: The line has a slope of 5, so its equation is (y 5x c), where (c) is the y-intercept we need to determine.

Step 1: Substitution

We substitute (y 5x c) into the circle's equation (x^2 y^2 81).

(x^2 (5x c)^2 81)

Step 2: Expanding and Simplifying

Expanding the equation:

(x^2 25x^2 10cx c^2 81)

(26x^2 10cx c^2 - 81 0)

Step 3: Using the Discriminant

For the line to touch the circle exactly once, the quadratic equation must have exactly one solution. This occurs when the discriminant of the quadratic equation is zero.

The discriminant (Delta) of (ax^2 bx c 0) is given by (Delta b^2 - 4ac).

(Delta (10c)^2 - 4(26)(c^2 - 81) 0)

(100c^2 - 104c^2 104 cdot 81 0)

(-4c^2 846 0)

(4c^2 846)

(c^2 211.5)

(c pm sqrt{211.5})

Conclusion with Two Equations

The line equations are:

(y 5x sqrt{211.5})

(y 5x - sqrt{211.5})

Geometric Method

We can also find the line equation using the geometric properties of the circle and the line.

Step 1: Finding the Gradient

The gradient of the line is given as 5. For the line to touch the circle at exactly one point, its gradient must be the same as the slope of the tangent at the point of contact.

Given the circle's equation (x^2 y^2 81), we differentiate it implicitly to find the slope of the tangent:

(2x 2yy' 0)

(y' -frac{x}{y})

The slope of the tangent is (-frac{x}{y}), and we are given (y' 5).

(-frac{x}{y} 5)

(x -5y)

Step 2: Substituting Back

We substitute (x -5y) into the circle's equation (x^2 y^2 81):

((-5y)^2 y^2 81)

(25y^2 y^2 81)

(26y^2 81)

(y^2 frac{81}{26})

(y pm frac{9}{sqrt{26}})

(x -5y mp frac{45}{sqrt{26}})

Step 3: Finding the Line Equation

We now have the points of contact (left(frac{45}{sqrt{26}}, -frac{9}{sqrt{26}}right)) and (left(-frac{45}{sqrt{26}}, frac{9}{sqrt{26}}right)).

The line equation through these points is (y 5x c). Substituting the point (left(frac{45}{sqrt{26}}, -frac{9}{sqrt{26}}right)):

(-frac{9}{sqrt{26}} 5left(frac{45}{sqrt{26}}right) c)

(-frac{9}{sqrt{26}} frac{225}{sqrt{26}} c)

(c -frac{234}{sqrt{26}})

(c -sqrt{234})

Therefore, the line equations are:

(y 5x - sqrt{234})

(y 5x sqrt{234})

Both methods yield the same equations, confirming our results.

Conclusion

In this article, we have explored how to find the equation of a line with a given slope and a specified distance from the origin. We have used both algebraic and geometric methods to reach the same conclusion. These techniques are not only useful in solving mathematical problems but also have practical applications in areas like computer graphics and engineering.