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

pentagon

step-by-step explanation:

the coordinates   a(5, -2), b(3,-1),c(-4,-4) d(-3,8) and e(-1,4) forms the vertices of the polygon when they are connected in order.

in order to classify that which polygon is it with the given coordinates, first count the number of vertices and then count the number of sides it can make with these vertices.

since, the given coordinates have five vertices, therefore, it will make five sides. moreover, pentagon is regular in nature therefore it will be convex.

now, a polygon with five sides is known as pentagon, therefore, the given polygon is the pentagon.