In the xy-plane, the graph of the equation x^2 - 8x + y^2 = 0 is a circle. What is the rad… — SAT

In the $xy$ -plane, the graph of the equation $x^2 - 8x + y^2 = 0$ is a circle. What is the radius of the circle?

Cevap: 4

The radius of the circle is 4.

By completing the square on the

x

-terms in the equation

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

, we add

( -8/2 )^2 = 16

to both sides, yielding

(x - 4)^2 + y^2 = 16

. Comparing this to the standard equation of a circle,

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

, we find that

r^2 = 16

. Taking the square root of both sides gives a radius of

4

Group the

x

-terms together and prepare to complete the square.

(x^2 - 8x) + y^2 = 0

To write the equation in the standard form of a circle,

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

, we need to complete the square for the quadratic expression in

x

Complete the square for the

x

terms by adding

16

to both sides of the equation.

(x^2 - 8x + 16) + y^2 = 16

, which simplifies to

(x - 4)^2 + y^2 = 16

Adding

( -8/2 )^2 = 16

to both sides creates a perfect square trinomial

(x-4)^2

on the left side of the equation.

Identify the value of

r^2

and find the radius

r

r^2 = 16

, which gives

r = \sqrt{16} = 4

Comparing the equation

(x - 4)^2 + y^2 = 16

to the standard form

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

shows that

r^2 = 16

. Since the radius must be positive,

r = 4

Completing the square to find the standard form equation of a circle,

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

, to determine its radius.

Soru