SOLUTION: Hi, i just need some help. Find the values of the constant k so that the graph of x^2+y^2+6x-4y=k^2-k-33 is a.) a circle, b.) a single point and c.) an empty set. I found a same pr

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Hi, i just need some help. Find the values of the constant k so that the graph of x^2+y^2+6x-4y=k^2-k-33 is a.) a circle, b.) a single point and c.) an empty set. I found a same pr      Log On


   

Question 1144028: Hi, i just need some help. Find the values of the constant k so that the graph of x^2+y^2+6x-4y=k^2-k-33 is a.) a circle, b.) a single point and c.) an empty set. I found a same problem but i need a whole solution thank you very much
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E2%2By%5E2%2B6x-4y=k%5E2-k-33

%28x%5E2%2B6x%29%2B%28y%5E2-4y%29=k%5E2-k-33
%28x%5E2%2B6x%2B3%5E2%29-3%5E2%2B%28y%5E2-4y%2B2%5E2%29-2%5E2=k%5E2-k-33
%28x%2B3%29%5E2-9%2B%28y-2%29%5E2-4=k%5E2-k-33
%28x%2B3%29%5E2%2B%28y-2%29%5E2-13=k%5E2-k-33
%28x%2B3%29%5E2%2B%28y-2%29%5E2=k%5E2-k-33%2B13
%28x%2B3%29%5E2%2B%28y-2%29%5E2=k%5E2-k-20
=> it’s a circle with center at (-3,2) and radius sqrt%28k%5E2-k-20%29
so, calculate k: to have real solution find for what value of k} is k%5E2-k-20=0+

k%5E2-k-20=0.....factor completely
k%5E2%2B4k-5k-20=0
%28k%5E2%2B4k%29-%285k%2B20%29=0
k%28k%2B4%29-5%28k%2B4%29=0
%28k-5%29%28k%2B4%29=0
solutions:
k=5 or+k=-4
using tese solution we will have radius equal to zero
then, the graph of x%5E2%2By%5E2%2B6x-4y=k%5E2-k-33 is
a.) a circle
if k%3E5+or k%3C-4
or k+is in interval: (-infinity, -4) and (5, infinity)
b.) a single point
if k=5 or k=-4
c.) an empty set
if k%3C5 and k%3E-4 or k+ is in interval -4%3Ck%3C5

Answer by ikleyn(52867) About Me  (Show Source):
You can put this solution on YOUR website!
.
You are given an equation of a circle in the "general form".

The first step is to transform it into the "standard form".



For it, perform completing the squares for x- and y-terms separately.

I will do it below, step by step. Trace my steps (!)


    x^2 + y^2 + 6x - 4y = k^2 - k - 33.                       (<<<---=== the original equation)


    (x^2 + 6x) + (y^2 - 4y) = k^2 - k - 33                    (<<<---=== I grouped / re-grouped the terms)

    (x^2 + 6x + 9) + (y^2 - 4y + 4) = k^2 - k - 33 + 9 + 4    (<<<---=== I added some constant terms in both sides)

    (x+3)^2        + (y-2)^2        = k^2 - k - 20            (<<<---=== I completed the squares in the left side 
                                                                         and combined like terms in the right side)


Now I have a "standard form" of the circle equation.


In order for this equation describes a real circle, its right side should be positive.

If the right side is zero, then the circle becomes a single point.

If the right side is negative, then the equation describes empty set.



So, if  k^2 - k - 20 > 0, then you have a circle with the center at the point (-3,2) in a coordinate plane (x,y);

    if  k^2 - k - 20 = 0, then the circle degenerates to the single point (-3,2);

    if  k^2 - k - 20 = 0, then the equation describes the empty set.



The parabola y = k^2 - k - 20 = (k-5)*(k+4)  has the roots k= 5 and k= -4.

    It is positive  if  k < -4  ot  k > 5;

    it is 0         if k= -4 or k= 5;

    it is negative  if  -4 < k < 5.



Therefore, the given (original) equation 

    - represents a circle,       if  k < -4  ot  k > 5;

    - represent a single point,  if  k= -4 or k= 5;

    - represents the empty set,  if  -4 < k < 5.

Solved, explained and completed.