Respuesta :
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.
[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