# Math problem about circles, graphs

1. Jul 28, 2008

### hancyu

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?

2. Jul 28, 2008

### slider142

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?

3. Jul 28, 2008

### hancyu

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.... whats that?

4. Jul 28, 2008

### hancyu

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?

5. Jul 28, 2008

### hancyu

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?

6. Jul 28, 2008

### spideyunlimit

(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

7. Jul 28, 2008

### spideyunlimit

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.

8. Jul 28, 2008

### spideyunlimit

and of course , the remaining values of k are for A CIRCLE

ie. k < -3.7445 and k > 7.7445

9. Jul 28, 2008

### spideyunlimit

hancyu, you are correct :)

10. Jul 28, 2008

Good work!