In the xy-plane, the equation 2x^2 + 2y^2 - 12x + 16y - 22 = 0 represents a circle. What i… — SAT

In the $xy$ -plane, the equation $2x^2 + 2y^2 - 12x + 16y - 22 = 0$ represents a circle. What is the diameter of this circle?

Cevap: 12

The diameter of the circle is 12.

Dividing the given equation

2x^2 + 2y^2 - 12x + 16y - 22 = 0

by 2 gives

x^2 + y^2 - 6x + 8y - 11 = 0

. Grouping the

x

and

y

terms and moving the constant to the right side gives

(x^2 - 6x) + (y^2 + 8y) = 11

. To complete the square, add

(\frac{-6}{2})^2 = 9

and

(\frac{8}{2})^2 = 16

to both sides, yielding

(x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16

, which simplifies to

(x-3)^2 + (y+4)^2 = 36

. Since the standard equation of a circle is

(x-h)^2 + (y-k)^2 = r^2

, the radius squared

r^2

is 36, which means the radius

r

is 6. The diameter is twice the radius, so

2 \times 6 = 12

Divide the entire equation by 2.

x^2 + y^2 - 6x + 8y - 11 = 0

To simplify the coefficients of

x^2

and

y^2

to 1, which is necessary before completing the square.

Group the variables and move the constant term.

(x^2 - 6x) + (y^2 + 8y) = 11

To isolate the quadratic and linear terms for both

x

and

y

on one side of the equation.

Complete the square for both variables by adding the appropriate values to both sides.

(x - 3)^2 + (y + 4)^2 = 36

Adding 9 (which is

(\frac{-6}{2})^2

) and 16 (which is

(\frac{8}{2})^2

) to both sides allows us to rewrite the trinomials as perfect squares:

(x-3)^2

and

(y+4)^2

. The right side becomes

11 + 9 + 16 = 36

Find the radius and calculate the diameter.

Diameter = 12

Comparing the equation to the standard form

(x-h)^2 + (y-k)^2 = r^2

shows that

r^2 = 36

, so the radius

r

is 6. The diameter is

2r = 2(6) = 12

Converting a circle's equation from general form to standard form by completing the square to identify its geometric properties.

Soru