A circle is described by the equation x2 + y2 + 14x + 2y + 14 = 0. What are the coordinates for the center of the circle and the length of the radius?
a. (-7, -1), 36 units
b. (7, 1), 36 units
c. (7, 1), 6 units
d. (-7, -1), 6 units

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naǫ
The equation of a circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h,k) - the coordinates of the center
r - the radius

[tex]x^2+y^2+14x+2y+14=0 \\ x^2+14x+49+y^2+2y+1+14-49-1=0 \\ (x+7)^2+(y+1)^2+14-49-1=0 \\ (x+7)^2+(y+1)^2-36=0 \\ (x+7)^2+(y+1)^2=36 \\ (x-(-7))^2+(y-(-1))^2=6^2 \\ \Downarrow \\ h=-7 \\ k=-1 \\ r=6[/tex]

The center is (-7,-1), the length of the radius is 6 units.
The answer is D.

Answer:

option d is correct.

center = (-7, -1)

r = 6

Step-by-step explanation:

The general equation of the circle is given by;

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

where;

(h, k) is the center of the circle

r is the radius of the circle respectively.

As per the statement:

A circle is described by the equation [tex]x^2 + y^2 + 14x + 2y + 14 = 0[/tex]

then;

[tex](x^2+14x)+(y^2 + 2y) + 14 = 0[/tex]

Using the completing the square:

[tex](x^2+14x+7^2-7^2)+(y^2 + 2y+1^2-1^2) + 14 = 0[/tex]

⇒[tex](x+7)^2 - 49 + (y+1)^2 - 1 + 14 = 0[/tex]

⇒ [tex](x+7)^2+(y+1)^2 -36 = 0[/tex]

Add 36 to both sides we have;

[tex](x+7)^2+(y+1)^2=36[/tex]

On comparing with the general equation of circle:

we get;

h = -7 ,  k = -1 and [tex]r^2 = 36[/tex]

⇒[tex]r = 6[/tex]

Therefore, the coordinate of the center of the circle is (-7, -1) and length of the radius of the circle is 6