Finding radius from circle equations in standard form

[MrNeWBiE]

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 ,,,

 

[rock.freak667]

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'.

 

[MrNeWBiE]

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 " ?

 

[rock.freak667]

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

 

[MrNeWBiE]

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 ,,,,

 

[nuketro0p3r]

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.

 

[MrNeWBiE]

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 ,,,

 

[Lord Dark]

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

 

[MrNeWBiE]

aha ,,,, thanks Lord

 

[Lord Dark]

no problem

 

[nuketro0p3r]

Would you please elaborate why?

 

[Lord Dark]

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)

 

[nuketro0p3r]

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 =)

 

[rock.freak667]

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.

 

[LCKurtz]

I don't understand why a circle can't have negative radius.

$(x-a)^2 + (y-b)^2 \ge 0$. You can't have it equal to a negative number and find any (x,y) that work.

 

[Mark44]

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.

 

[nuketro0p3r]

@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 =)