Math problem about circles, graphs

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    Circles Graphs
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Discussion Overview

The discussion revolves around determining the values of k for which the equation x^2 + y^2 + 2x - 4y + 26 = k^2 - 4k represents a circle, a point, or an empty set. The scope includes mathematical reasoning and problem-solving related to the properties of conic sections.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Exploratory

Main Points Raised

  • One participant reformulates the equation into the standard form of a circle and seeks guidance on the next steps.
  • Another participant prompts consideration of conditions that lead to no solutions or a single solution, questioning the possibility of the left side being negative.
  • A participant identifies that a negative radius indicates an empty set and attempts to solve for k using the quadratic equation.
  • Further contributions clarify that for the equation to represent a point, the right-hand side must equal zero, leading to specific k values.
  • Another participant provides conditions for the empty set and identifies the corresponding interval for k.
  • Participants discuss the remaining values of k that would indicate a circle, suggesting ranges based on previous findings.
  • Multiple participants express agreement on the derived values and conditions for k.

Areas of Agreement / Disagreement

Participants generally agree on the derived values of k and the conditions for the equation to represent a circle, a point, or an empty set. However, the discussion includes some uncertainty regarding the precise interpretation of the conditions and the intervals for k.

Contextual Notes

Some participants express uncertainty about the implications of the derived values and the conditions for different cases, indicating a need for further clarification on the relationships between k and the nature of the solutions.

hancyu
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this is the question...

For what values of k is the graph of the equation x^2 + y^2 +2x-4y+26=k^2 - 4k

a. a circle
b. a point
c. an empty set

i think i should change the equation to this form...(x+1)^2+(y-2)^2=k^2 - 4k - 29

then what?
 
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You already got your equation into the form of the equation of a circle. What would cause it to have no solutions? What would cause it to have only 1 solution?
Hint:Can the left side ever be negative?
 
i know that if r is negative...its an empty set.but how do i get k?

if i input k^2 - 4k - 29 = 0 in the calculator... k = 7.74 & -3.74... what's that?
 
r = 0 so that the eqn would become a pt.
then k = 7.74 & -3.74

how then do i know which numbers would make the eqn a circle/ empty soln?
 
oh! i think i got it!

a. -3.74>k or k>7.74
b. k = 7.74 or k=-3.74
c. -3.74>k<7.74

is this right?
 
(x+1)^2+(y-2)^2 = k^2 - 4k - 29

For empty set, make the right hand side quadratic equation negative. For a point make the right hand side turn out as 0 and for a circle make it turn out as anything else.

FOR A POINT ---> k^2 - 4k - 29 = 0
k = 4 +- sqrt (16 + 116) / 2 = 4+- rt (132) / 2 = 4 +- 11.489 / 2 =
7.7445 or -3.7445
 
FOR EMPTY SET :
k^2 - 4k - 29 < 0
(k - 7.7445)( k + 3.7445 ) < 0
k E (-3.7445 , 7.7445) ... i have given the interval for k.
 
and of course , the remaining values of k are for A CIRCLE

ie. k < -3.7445 and k > 7.7445
 
hancyu, you are correct :)
 
  • #10
hancyu said:
oh! i think i got it!

a. -3.74>k or k>7.74
b. k = 7.74 or k=-3.74
c. -3.74>k<7.74

is this right?

Good work!
 

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