Showing Tangency of a Line and Circle: Simplifying the Equation Method

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Homework Help Overview

The discussion revolves around determining whether a given line is tangent to a circle using their respective equations. The original poster explores algebraic methods to find points of intersection and the conditions for tangency.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss rearranging equations to find intersections, questioning the effectiveness of algebraic manipulation. There is mention of using the discriminant of a quadratic equation to establish tangency.

Discussion Status

Some participants provide alternative methods for proving tangency, including checking the length of the perpendicular from the center of the circle to the line. The original poster seeks validation for their approach, while others clarify the necessity of certain steps in the reasoning process.

Contextual Notes

There is a focus on ensuring that the line intersects the circle at only one point, which is a key aspect of tangency. Participants also reference specific equations and their transformations, indicating a reliance on algebraic techniques.

Gaz031
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If i have the equation of a circle and the equation of a line how do i show if the line is a tangent of the circle?

At first i tried rearranging them both in the form f(x) = 0, setting them to equal each other. Rearranging and simplifying, then eliminating x or y by using the linear line equation. I was then going to try to find the points at which they meet and hence differentiate both equations to find dy/dx at this point - if they were equal i would have shown it was a tangent.
However, this usually doesn't work. I'm not sure if there's something wrong with my algebraic manipulation but it's rarely solvable with the quadratic equation.

Is there an easier way? If so, what is it?
 
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Length of perendicular from centre of circle should be qual to radius of circle for the line to be tanget.
other way ut the equation of circle in form of one variable by using equation of line and show discriminant is zero of quadratic equation so formed. rovig line intersects at only one point.
 
Thanks for the help, would this be an accepable method?

Prove that x^2 + y^2 - 10y = 0 has a tangent 4x - 3y + 40 = 0.

Rearrange to form 4x = 3y-40, x = 3/4 y -10

Substitute into the equation of the circle:

(3/4 y - 10)^2 + y^2 - 10y = 0
Simplifies to y^2 -16y + 64 = 0

Solves to give y = only 8, hence at this point x = -4.

If x^2 + y^2 - 10y = 0
dy/dx = -2x/(2y-10)
Substitute in the x and y gives dy/dx = 8/6 = 4/3

Of the equation of the tangent line:
4x = 3y - 40
dy/dx = 4/3.

They meet at only one point and at this point they have the same gradient, hence they are tangents.
 
Gaz,
When u say,
"...Substitute into the equation of the circle..." (fourth line)

What are u doing here ??...
u are finding the intersection points of the line with the circle.

Now every line usually will intersect in two points with the circle but our given line intersects at one and only one point.

So by definition of tangent, the given line is tangent to the circle.

So the following working,
"If x^2 + y^2 - 10y = 0
dy/dx = -2x/(2y-10)
Substitute in the x and y gives dy/dx = 8/6 = 4/3

Of the equation of the tangent line:
4x = 3y - 40
dy/dx = 4/3."

is not necessary.

Essentially, u have solved the problem in complete. :)

-- AI
 

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