How to Find the Center and Radius of a Circle: Circle Equation Help

  • Thread starter MrNeWBiE
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In summary, x^2 + y^2 + 4x + 8y + 25 =0 is not a circle equation because (x+2)^2+(y+4)^2=-20 and the radius is minus.
  • #1
MrNeWBiE
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hi

Homework Statement



determine if whether it's circle or not , if yes find the center and radius

1) 2x^2 + 2y^2 - 5x + 4y - 1 =0

2) x^2 + y^2 + 4x + 8y + 25 =0



The Attempt at a Solution



1) center (5/4,-1)
2) center (2,4)

well i need help in how to find the " radius " and how to know if it's circle equation or not ,,,
 
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  • #2
If it is a circle, it can be put into the form

(x-a)2+(y-b)2=r2

with center (a,b) and radius 'r'.
 
  • #3
i know that ,,,, but now it's expanded ,,,

so i will got b^2 + a^2 - r^2 in one side ,,,,,

how i will know if it's circle and the " r " ?
 
  • #4
MrNeWBiE said:
i know that ,,,, but now it's expanded ,,,

so i will got b^2 + a^2 - r^2 in one side ,,,,,

how i will know if it's circle and the " r " ?

Gather all the terms with 'x' and compete the square, do the same for the 'y', then move all the constants to the right side of the equation and that will be your r2
 
  • #5
i completed the square and i got the points ,,, but how to know if it's circle equation ? because #2 is not ,,,
but i don't know why ,,,,
 
  • #6
The general equation of circle are in the form of x^2+y^2+2gx+2fy+c=0 and since the equations that you mentioned are in this form I am pretty sure they're a circle.
 
  • #7
2) x^2 + y^2 + 4x + 8y + 25 =0
is not circle equation am sure because I see it in my book ,,,,
and I don't know why it's not ,,,
 
  • #8
No it's not a circle ,, Put "x^2 + y^2 + 4x + 8y + 25 =0 " in circle equation :
(x+2)^2+(y+4)^2=-20 ,, the radius is minus so it's not a circle :smile:
 
  • #9
aha ,,,, thanks Lord
 
  • #10
no problem :biggrin:
 
  • #11
Would you please elaborate why?
 
  • #12
nuketro0p3r said:
Would you please elaborate why?

Why what ? it's not a circle ?? ,, if yes here's why :
the equation
x^2 + y^2 + 4x + 8y + 25 =0
>> (x^2+4x+...)+(y^2+8y+...) = -25 [complete the square for both of them]
>> (x+2)^2+(y+4)^2 = -25+4+16 [add on both sides 4 & 16 to be able to complete the square]
>>(x+2)^2+(y+4)^2= -5 [ you get the radius -5 which tells you that there is no circle with minus radius]

hope you understand now ,, and sorry for the wrong answer before (r=-20)
 
  • #13
I don't understand why a circle can't have negative radius. For example, consider this circle with equation x^2+y^2=0, which has its center at origin and passes through points P(x, 0) & Q(-x, 0). If the radius of a circle is defined as the distance from the center to any point on the circle, then the circle in the above example would have negative radius in the 2nd and 3rd quadrant. Thanks for the reply in advance =)
 
  • #14
nuketro0p3r said:
I don't understand why a circle can't have negative radius. For example, consider this circle with equation x^2+y^2=0, which has its center at origin and passes through points P(x, 0) & Q(-x, 0). If the radius of a circle is defined as the distance from the center to any point on the circle, then the circle in the above example would have negative radius in the 2nd and 3rd quadrant. Thanks for the reply in advance =)

Distances aren't negative.
 
  • #15
nuketro0p3r said:
I don't understand why a circle can't have negative radius.

[itex](x-a)^2 + (y-b)^2 \ge 0[/itex]. You can't have it equal to a negative number and find any (x,y) that work.
 
  • #16
nuketro0p3r said:
I don't understand why a circle can't have negative radius. For example, consider this circle with equation x^2+y^2=0, which has its center at origin and passes through points P(x, 0) & Q(-x, 0). If the radius of a circle is defined as the distance from the center to any point on the circle, then the circle in the above example would have negative radius in the 2nd and 3rd quadrant. Thanks for the reply in advance =)
The equation x^2 + y^2 = 0 represents a degenerate circle, a circle whose radius is 0. This "circle" DOES NOT pass through points P(x, 0) and Q(0, y), unless x = 0 and y = 0.
 
  • #17
@rock.freak667: I know but I got a little confused by the structure of the equation.
@LCKurtz: Thanks that helps me understand the concept better :)
@Mark44: I actually though about that after posting :D

Thanks for the help guys =)
 

1. What is the equation for finding the center and radius of a circle?

The equation for finding the center and radius of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) represents the coordinates of the center and r represents the radius.

2. How do you find the center of a circle given its equation?

To find the center of a circle given its equation, set the equation equal to the standard form (x - h)^2 + (y - k)^2 = r^2 and solve for (h,k). The values of (h,k) will represent the coordinates of the center of the circle.

3. How do you find the radius of a circle given its equation?

To find the radius of a circle given its equation, take the square root of both sides of the equation (x - h)^2 + (y - k)^2 = r^2. The value of r will represent the radius of the circle.

4. Can you find the center and radius of a circle if you only know three points on its circumference?

Yes, you can find the center and radius of a circle if you only know three points on its circumference. Use the formula (x - h)^2 + (y - k)^2 = r^2 and plug in the coordinates of the three points to create a system of equations. Then, solve the system of equations to find the values of (h,k) and r.

5. Is there a different method for finding the center and radius of a circle if the equation is in standard form?

No, the method for finding the center and radius of a circle is the same whether the equation is in standard form or not. The standard form (x - h)^2 + (y - k)^2 = r^2 is just a more organized way to write the equation.

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