Question

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?

Answer 1

Hello,
x²+2*7x+49+y²+2y+1+14-49-1=0
==>(x+7)²+(y+1)²=36

Center is (-7,-1) and radius 6.

Answer 2

Answer:  The co-ordinates of the center of the circle are (-7, -1) and the length of the radius is 6 units.

Step-by-step explanation:  Given that a circle is described by the following equation :

$x^2+y^2+14x+2y+14=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We are to find the co-ordinates for the center and the length of the radius of the given circle.

The STANDARD equation of a circle with center (h, k) and radius of length r units is given by

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

From equation (i), we have

$x^2+y^2+14x+2y+14=0\\\\\Rightarrow (x^2+14x)+(y^2+2y)+14=0\\\\\Rightarrow (x^2+2\times x\times 7+7^2)+(y^2+2\times y\times1+1^2)+14-7^2-1^2=0\\\\\Rightarrow (x+7)^2+(y+1)^2+14-49-1=0\\\\\Rightarrow (x+7)^2+(y+1)^2-36=0\\\\\Rightarrow (x+7)^2+(y+1)^2=36\\\\\Rightarrow (x-(-7))^2+(y-(-1))^2=6^2.$

Comparing with the standard form, we get

center, (h, k) = (-7, -1)  and  radius, r = 6 units.

Thus, the co-ordinates of the center of the circle are (-7, -1) and the length of the radius is 6 units.