Finding the center and radius of circles given an equation

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The equation of the circle x² + y² + 4y - 117 = 0 was transformed into standard form as x² + (y + 2)² = 121. In this form, the center of the circle is identified as the point (0, -2), derived from the equation format (x - h)² + (y - k)² = r². The radius is calculated as 11, which is the square root of 121. This confirms the correct identification of both the center and radius of the circle.
Jim4592
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Homework Statement


Find the center and the radius of the circle with the given equation.


Homework Equations



x2+y2+4y-117=0

The Attempt at a Solution



I first got it in standard form by completing the square:

x2+(y+2)2=121

but i don't know how to get the center and radius of it? I'm pretty sure its given in that form of the equation, right? but I'm not sure which numbers are the ordered pair.
 
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Jim4592 said:

Homework Statement


Find the center and the radius of the circle with the given equation.


Homework Equations



x2+y2+4y-117=0

The Attempt at a Solution



I first got it in standard form by completing the square:

x2+(y+2)2=121

but i don't know how to get the center and radius of it? I'm pretty sure its given in that form of the equation, right? but I'm not sure which numbers are the ordered pair.

Your basic standard form for a circle equation is (x-h)2+(y-k)2=r2. That form let's you directly read how the center has been translated and also shows the radius.
 
The general equation for a circle is:

(x + a)² + (y + b)² = r²

where (-a,-b) is the centre of the circle and r is the radius.

That should complete your problem, as you have already put it mostly into the right form
 
so the center of my circle is (0,-2) and would that make the radius 11, the square root of 121??
 
Jim4592 said:
so the center of my circle is (0,-2) and would that make the radius 11, the square root of 121??

(hmm … everyone else seems to have gone out, so …)

Yup! :biggrin:
 

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