What Are the Possible Values of m for a Tangent Line to a Circle?

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Homework Help Overview

The discussion revolves around finding the possible values of the slope \( m \) for a line tangent to a given circle in the coordinate plane. The original poster is exploring the relationship between the line and the circle defined by their respective equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to substitute the line equation into the circle's equation but encounters difficulties with multiple unknowns. Some participants suggest rearranging the resulting equation into a standard quadratic form and applying the quadratic formula.

Discussion Status

Participants are actively engaging with the problem, offering guidance on how to manipulate the equations and check conditions for tangency. There is a recognition of the need to ensure the discriminant is non-negative for the quadratic equation derived from the substitution.

Contextual Notes

There is an emphasis on understanding the conditions for tangency, specifically relating to the discriminant of the quadratic equation formed. The original poster expresses a lack of experience with certain algebraic manipulations, indicating a learning context.

Gaz031
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I'm just introducing myself to coordinate geometry in the xy plane of cirlces.
Here's a question I'm having trouble with:

Q11: The line with equation y=mx is a tangent to the circle with equation x^2 + y^2 - 6x - 6y + 17 = 0. Find the possible values of m.

At first i thought i'd try substituting y=mx into the curve equation, but i was still left with 2 unknowns. I don't really know what to do here. Could anyone offer some advice? Thanks.
The answer is (9 (+ or -) root 17) / 8
 
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After substituting you should have

[tex]x^2-m^2x^2-6x-6mx+17=0[/tex]

Rearrange this and you get

[tex](1+m^2)x^2 - (6m+6)x + 17 = 0[/tex]

Let a = 1 + m2, b = -(6m + 6) and c = 17. Then use the quadratic equation (and remember that you want the expression under the square root to be greater than or equal to 0).
 
Last edited:
Your latex looks ok. Do you mean rearrange to form:
x^2 ( 1+ m^2) - x(6m+1) + 17 = 0 ?

Edit: Ah yes you typed that below. I'll try that now.
 
I re-edited my post. Please have a look.
 
I get it. So you say that b^2 - 4ac = 0. Then form yet another quadratic equation, tidy up and simplify. I haven't had much experience of using brackets as coefficients in the quadratic equation but that's something I'm going to remember for the future.

Thanks for the help!
 
That should be b2 - 4ac ≥ 0.
 

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