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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)?

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apologiabiology

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

[tex](x-h)^2+(y-k)^2=r^2[/tex]

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

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

[tex]D=\sqrt{(-4-(-2))^2+(0-1)^2}[/tex]

[tex]D=\sqrt{(-4+2)^2+(-1)^2}[/tex]

[tex]D=\sqrt{(-2)^2+1}[/tex]

[tex]D=\sqrt{4+1}[/tex]

[tex]D=\sqrt{5}[/tex]

D=r

so the equation is

[tex](x-(-4))^2+(y-0)^2=(\sqrt{5})^2[/tex]

or

[tex](x+4)^2+y^2=5[/tex]

raphealnwobi

The standard equation of a circle is given by:

(x – h)² + (y – k)² = r²

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

Given that the circle has center at (–4, 0) and passes through the point (–2, 1). The radius (r) is the distance between (–4, 0) and (–2, 1). Hence:

[tex]r=\sqrt{(-2-(-4))^2+(1-0)^2}=\sqrt{5}[/tex]

(x – (-4))² + (y – 0)² = (√5)²

(x + 4)² + y² = 5

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