For what value of k will (k – 5)x²+ 2ky²-5x + 6y – 3 = 0 represents a circle? (

This isn't going to be a circle even if we choose k to be -5.

I think there has to be an error in the question.

Step-by-step explanation:

We want the coefficient of x^2 and y^2 to be the same so we will want to solve:

k-5=2k

Subtract k on both sides:

-5=k

You should verify it will give a circle.

(k – 5)x²+ 2ky²-5x + 6y – 3 = 0 with k=-5:

(-5– 5)x²+ 2*-5y²-5x + 6y – 3 = 0

-10x²+ -10y²-5x + 6y – 3 = 0

Add 3 on both sides:

-10x²+ -10y²-5x + 6y = 3

Divide both sides by -10:

x²+ y²+5/10x -6/10y = -3/10

Gather the x's together and gather the y's together:

x²+5/10x+y²-6/10y = -3/10

We will begin to take steps to complete square for each the x part and the y part.

For each v^2+kv, add (k/2)^2 with goal to write as (x+k/2)^2. Remember whatever you do to one side you must do to the other.

x²+5/10x+(5/[2×10])^2+y²-6/10y+(-6/[2×10])^2 = -3/10+(5/[2×10])^2+(-6/[2×10])^2

Since radius square is not positive we can stop here. The radius has to be a real number after all. If you aren't convinced you can continue to write in center-radius form.

(x+1/4)²+(y-3/10)² = -3/10+1/16+9/100

(x+1/4)²+(y-3/10)² = -3/10+1/16+9/100

(x+1/4)²+(y-3/10)² = -59/400