Question

What is the equation of a circle with a center at (-4,0) that passes through the point (-2,1)

Answer 1

Hello from MrBillDoesMath!

Answer:

The third choice, (x+4)^2 + y^2 = 5

Discussion:

Since the center is (-4.0) the equation of the circle with radius "r" is:

(x - (-4))^2 + ( y - 0)^2 = r^2 => as -(-4) = 4

(x+4)^2 + y^2 = r^2

Look at the choices provided. Only the third choice is of this form so the answer is

(x+4)^2 + y^2 = 5 (*)

Note this circle passes through (-2,1) as

(-2 + 4)^2 + (1)^2 =

(-2)^2 + 1^2 =

4 + 1 = 5

which agrees with (*)

Thank you,

MrB

Answer 2

The equation of a circle with a center at (-4,0) that passes through the point (-2,1) is x² + y² + 2x - 2y =0

Explanation:

It is given that,

a circle with center at (-4,0) that passes through the point (-2,1)

To find the radius of circle.

Radius r = √[(-2 --4)² + (1-0)²]

r = √[(-2 +4)² + 1²]

= √[2² + 1²] = √(4 + 1) =√5

To find the equation of circle

(x - -2)² + (y - 1)² = (√5)²

(x + 2)² + (y - 1)² = 5

x² + 2x + 4 + y² -2y +1 = 5

x² + y² + 2x - 2y + 5 = 5

x² + y² + 2x - 2y =0