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

Click For Summary
SUMMARY

The discussion focuses on determining the possible values of m for the line equation y=mx to be tangent to the circle defined by x² + y² - 6x - 6y + 17 = 0. The correct approach involves substituting y=mx into the circle's equation, leading to the quadratic form (1+m²)x² - (6m+6)x + 17 = 0. The solution requires applying the quadratic formula and ensuring the discriminant (b² - 4ac) is greater than or equal to zero, resulting in the values of m as (9 ± √17) / 8.

PREREQUISITES
  • Understanding of coordinate geometry and tangents to circles
  • Familiarity with quadratic equations and the quadratic formula
  • Basic algebraic manipulation skills
  • Knowledge of discriminants in quadratic equations
NEXT STEPS
  • Study the properties of tangents to circles in coordinate geometry
  • Learn about the quadratic formula and its applications
  • Explore the concept of discriminants and their significance in determining the nature of roots
  • Practice solving quadratic equations with variable coefficients
USEFUL FOR

Students learning coordinate geometry, mathematics educators, and anyone interested in solving problems involving tangents to circles and quadratic equations.

Gaz031
Messages
51
Reaction score
0
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
 
Physics news on Phys.org
After substituting you should have

x^2-m^2x^2-6x-6mx+17=0

Rearrange this and you get

(1+m^2)x^2 - (6m+6)x + 17 = 0

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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
Replies
3
Views
2K
Replies
8
Views
3K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 14 ·
Replies
14
Views
2K