Question

Determine the equation of a circle with a center at (–4, 0) that passes through the point (–2, 1) by following the steps below. Use the distance formula to determine the radius: Substitute the known values into the standard form: (x – h)² + (y – k)² = r². What is the equation of a circle with a center at (–4, 0) that passes through the point (–2, 1)?

Answer 1

Answer:

(x + 4)² + y² = 5

Step-by-step explanation:

Standard equation of the circle is,

(x - h)² + (y - k)² = r²

Where (h, k) is the center and r is the radius of the circle.

One point on the circle has been given as (-4, 0) and point on the circle is (-2, 1).

Distance between center and the point on the circle = Radius

Distance between two points = $\sqrt{(x_{2}-x_{2})^{2}+(y_{2}-y_{2})^{2}}$

Now distance between (-2, 1) and (-4, 0) will be,

r = $\sqrt{(-4+2)^{2}+(0-1)^{2}}$

r = $\sqrt{(-2^{2}+(-1)^{2})}$

r = $\sqrt{5}$

Now we substitute the values in the standard equation of the circle,

(x + 4)² + (y - 0)²= (√5)²

(x + 4)² + y² = 5

Answer 2

the equation of a line that has a center at (h,k) and radius of r is

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

we can use the distance formula to find the radius ( I would just substitute but whatever)

distance between (-4,0) and (-2,1) is

$D=\sqrt{(-4-(-2))^2+(0-1)^2}$

$D=\sqrt{(-4+2)^2+(-1)^2}$

$D=\sqrt{(-2)^2+1}$

$D=\sqrt{4+1}$

$D=\sqrt{5}$

D=r

so the equation is

$(x-(-4))^2+(y-0)^2=(\sqrt{5})^2$

or

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