Math

posted by Shenaya

The equation 2x^2+2y^2+x+4y+2c=0 represents a circle.(let's name it S1)

1)Find the values of a,b and and show that c<(17/16)

2)For positive integers of c,if the above circle is divided into two by, (x+p)^2=p^2 circle,find the value of the constant p.


1)I know that the coefficients of x^2 and y^2 in an equation of a circle should be equal. So a=2

And an equation of a circle shouldn't have terms of (xy),so b=0

And if we compare this with a standard equation of a circle(x^2+y^2+2gx+2fy+c=0)
the radius of the circle(let it be r) r=√((-g)^2+(-f)^2-c)

So for the above equation to represent a real circle r>0

So g^2+f^2-c>0
c<g^2+f^2

and in this case,
2c<(-1/2)^2+(-2)^2
2c<(1/4)+(4)
2c<(17/4)
c<(17/8)

I can't find where I made the mistake.

2)I think for this one we should substract the left sides of the equations and then we get another equation which is equal to zero.

But how do we find p?
How do we use the given data,to find it's value?

  1. Steve

    I think you lost a factor of 2 in your transformation.

    2x^2+2y^2+x+4y+2c=0
    x^2 + x/2 + y^2+2y = -c
    (x + 1/4)^2 + (y+1)^2 = -c + 1/16 + 1
    For this to be a real circle, the right side must be positive, so
    c < 17/16


    The only positive integer value of c is 1, so we have
    (x + 1/4)^2 + (y+1)^2 = 1/16
    Now, we also have
    (x+p)^2 = p^2
    x+p = ±p
    x = 0 or -2p
    So, we have the circle with center at (-1/4,-1) cut in half by the line x = -2p. That line goes through the center, so
    -2p = -1/4
    p = 1/8

    see

    http://www.wolframalpha.com/input/?i=plot+(x+%2B+1%2F4)%5E2+%2B+(y%2B1)%5E2+%3D+1%2F16,+(x%2B1%2F8)%5E2+%3D+1%2F64

  2. Shenaya

    Thank you!

  3. Shenaya

    And what if it's (x+p)^2 + y^2 =p^2 ,instead of (x+p)^2 = p^2 ? How do we eliminate y^2 ?

  4. Steve

    that's a whole different problem. You want to have two circles intersect so that they divide in half? Not nearly so easy, as I'm sure you can appreciate.

    (x + 1/4)^2 + (y+1)^2 = 1/16
    (x+p)^2 + y^2 = p^2

    x^2 + 1/2 x + 1/16 + y^2 + 2y + 1 = 1/16
    x^2 + 2px + p^2 + y^2 = p^2

    x^2 + 1/2 x + y^2 + 2y = -1
    x^2 + 2px + y^2 = 0
    Now if you subtract you get
    (1/2 - 2p)x + 2y = -1
    y = (p - 1/4)x - 1/2

    If p=1, the two circles just touch at (-2/5,-4/5)

    For p < 1, the two circles do not intersect. So, we need p > 1.

    Now if you want the 1st circle to be cut in half, you have to integrate. If you haven't yet studied calculus, that will be beyond the scope of this article ...

  5. Steve

    Actually, this is just a problem in geometry. I would not be surprised if Euclid did it 2000 years ago. Still, you could start here:

    http://mathworld.wolfram.com/Circle-CircleIntersection.html

    But using general circles in coordinate geometry, it will get messy!

  6. Shenaya

    I've studied calculus. And we're asked to find this for positive integers of c,(c<(17/16) )

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