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How do you solve this? Write the standard form of the equation of a circle with center...

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zigtozag | Student, College Freshman | eNotes Newbie

Posted January 14, 2009 at 2:55 AM via web

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How do you solve this? Write the standard form of the equation of a circle with center (0, 0), given r = 8.

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giorgiana1976 | College Teacher | Valedictorian

Posted January 14, 2009 at 3:50 AM (Answer #1)

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The equation of a circle with the center in M(a,b)=O(0,0) and r=8

(x-a)^2+(y-b)^2=r^2

(x-0)^2+(y-0)^2=64 x^2+y^2=64 
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neela | High School Teacher | Valedictorian

Posted July 9, 2009 at 10:08 AM (Answer #2)

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The standard form of a circle's equation with  (-g,-f) as centre is :

x^2+y^2+2gx+2fy+c=0

The equation of the given circle with centre (0,0) and with radius  8 is (x-0)^2+(y-0)^2 =8^2.

Rearranging the equation, we get:

x^2+ y^2+2*0*x+2*0*y+(-64)=0, in standard form with g=0, f=0 and c=-64.

The simplified form is: x^2+y^2-64=0.

Hope this helps.

 

 

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revolution | College Teacher | Valedictorian

Posted July 18, 2010 at 11:26 PM (Answer #3)

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The general formula for equation of circles is:

(x-a)^2+(y-b)^2=r^2, in the standard form, where (a,b) is the centre and r is the radius.

Since centre is (0,0) and raidus=8

Sub numerals into equation

(x-0)^2 + (y-0)^2= (8)^2

x^2+y^2=64

 

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atyourservice | TA , Grade 10 | Valedictorian

Posted July 30, 2014 at 9:14 AM (Answer #5)

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(x-a)^2+(y-b)^2=r^2

centre is (0,0) and radius=8

x=0 y=0

plug it in

(x-0)^2+(y-0)^2=8^2

simplify:

x^2+y^2=64 or x^2+y^2-64=0.

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givingiswinning | Student, Grade 10 | Valedictorian

Posted August 15, 2014 at 11:11 PM (Answer #6)

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x=0 y=0

(x-0)^2+(y-0)^2=8^2

simplify

x^2+y^2=64 or x^2+y^2-64=0.

 
 
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vedhajothi | Student, Undergraduate | eNotes Newbie

Posted July 5, 2009 at 1:37 AM (Answer #4)

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the equation of the circle with centre at the origin is given by

x^2+y^2=r^2

therefore, our equation becomes

x^2+y^2=64(since r=8)

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