SOLUTION: what is the value of k in x^2+y^2-6x+8y+5=k in a circle, a point, and an empty set? please answer.

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Question 1008349: what is the value of k in x^2+y^2-6x+8y+5=k in a circle, a point, and an empty set? please answer.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Complete the square.
x%5E2-6x%2By%5E2%2B8y=k-5
%28x%5E2-6x%2B9%29%2B%28y%5E2%2B8y%2B16%29-9-16=k-5
%28x-3%29%5E2%2B%28y%2B4%29%5E2=k%2B20
If the right hand side is greater than zero, then it is the square of the radius of the circle centered at (3,-4).
k%2B20%3E0
k%3E-20
.
.
If the right hand side is equal to zero, then it is a single point, namely the center (3,-4).
k%2B20=0
k=-20
.
.
If the right hand side is less than zero, then the solution is the empty set.
k%3C-20