# A circle has the equation x2+10x+y2+9=0. What is its center?

Updated 6 days ago:

x^2+10x+y^2+9=0.

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• x^2+10x+y^2+9=0

=>

x^2+10x+(10/2)^2-(10/2)^2+y^2+9=0

=>

(x+5)^2+(y+0)^2-25+9=0

=>

(x+5)^2+(y+0)^2=16

=>

the center of this circle is (-5,0)

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• do complete the square for x and for y

(x^2 +10x + 25) + y^2 = 25 - 9

==> (x+5)^2 +y^2 = 16

so centre is (-5,0) and radius 4 units.

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• x^2 +10x +y^2 + 9 = 0. Completing the square gives x^2+10x+25 + y^2 = 4^2, ie.,

(x+5)^2 + (y-0)^2 = 4^2, which defines a circle, radius 4, center (-5,0).

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• A  circle has the equation x^2 + 10x + y^2 + 9 = 0.

circle

center | (-5, 0)

diameter | 8

area enclosed | 16 π≈50.2655

circumference | 8 π≈25.1327

centroid | (-5, 0)

curvature | 1/4 = 0.25

Its center is (-5, 0)

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• Complete the Square in each letter

(x + 5)^2 - (5)^2 =>

(x + 5)^2 = 25

Hence x = -5

Since there are no terms in 'y' then y = 0

Hence centre is (-5,0)

-9 + 25 = 16

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• x² + 10x + y² + 9 = 0

so, (x + 5)² + y² + 9 = 25

Hence, (x + 5)² + y² = 16 => 4²

i.e. centre (-5, 0) and radius 4

A sketch is below.

:)> • Wayne DeguMan
Lv 7
6 days agoReport

If you want to look through my 17,000+ answers, you can decide. Thanks anyway.

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• Any circle in the plan can be written in this general form:

x² + y² + Dx + Ey + F = 0

If it is in fact a circle, its radius is (-D/2, -E/2).

In your case, D = 10, E = 0, so (-5, 0) is the center.

You should also check the radius though.

radius = √(D² + E² - 4F)/2

If D² + E² - 4F < 0, the equation has no solutions in the plane.

If D² + E² - 4F = 0, then the locus is only a single point. See the center.

If D² + E² - 4F > 0, then yes, it is a circle.

D² + E² - 4F = (10)² + (0)² - 4(9) = 64

It is a circle with center(-5, 0) and radius 4.

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• x^2+10x+y^2+9=0

x^2 + 10x + y^2 = - 9

(x^2 + 10x + 25) + y^2 = - 9 + 25

(x + 5)^2 + y^2 =√(16)

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

the center of the circle is (-5,0) answer//

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• compete the square .. twice

x^2+10x+y^2+9=0.

(x + 5)^2  +  (y + 0)^2 = 16

center = (-5, 0)

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