- #1

Jaco Viljoen

- 160

- 9

**Mod note**: Moved from technical math section, so no template.

Hi All, i couldn't find any geometry under coursework and homework, but saw some geometry under general math.

I hope this is the right place, please advise and move if I have posted in the wrong place.The circle with equation x

^{2}+y

^{2}-2x+4y=0 has a centre C and radius r where

1. C(1,-2); r=Sqrt 5

2. C(-1,2); r=Sqrt 5

3. C(1,-4); r=sqrt 17

4. C(-1,4); r=17

5. C(1,-2); r=5

I have tried a couple of things:

Simplifying the equation:

x

^{2}+y

^{2}-2x+4y=0

x(x-2)+y(y+4)

x=2 and y=-4 This doesn't correspond to any of my options so can't be correct?

Using the standard form of the equation of a circle:

1.

(x-h)

^{2}+(y-k)

^{2}=r

^{2}

(x-1)

^{2}+(y+2)

^{2}=r

^{2}

x

^{2}-x-x+1+y

^{2}+2y+2y+4=r

^{2}

x

^{2}-2x+1+y

^{2}+4y+4=r

x

^{2}+y

^{2}-2x+4y+5=r

^{2}

x

^{2}+y

^{2}-2x+4y=-5

radius can't be negative...

so r=sqrt5

2.

(x-h)

^{2}+(y-k)

^{2}=r

^{2}

(x+1)

^{2}+(y-2)

^{2}=r

^{2}

(x+1)(x+1)+(y-2)(y-2)=r

^{2}

x

^{2}+1x+1x+1+y

^{2}-2y-2y-4=r

^{2}

x

^{2}+2x+y

^{2}-4y-3=r

^{2}

x

^{2}+y

^{2}+2x-4y=sqrt3

This doesn't correspond to any of my options so can't be correct.

3.

(x-h)

^{2}+(y-k)

^{2}=r

^{2}

(x-1)

^{2}+(y-(-4))

^{2}=2

^{2}

(x-1)(x-1)+(y-2)(y-2)=r

^{2}

x

^{2}-1x-1x+1+y

^{2}-2y-2y+4=r

x

^{2}+y

^{2}-2x-4y+5=r

^{2}

doesn't match the circle's equation

4.

(x-h)

^{2}+(y-k)

^{2}=r

^{2}

(x+1)

^{2}+(y-4)

^{2}=r

^{2}

(x+1)(x+1)+(y-4)(y-4)=r

^{2}

x

^{2}+1x+1x+1+y

^{2}-4y-4y+8=r

^{2}

x

^{2}+y

^{2}+2x-8y+9=r

^{2}

doesn't match the circle's equation

5.

(x-h)

^{2}+(y-k)

^{2}=r

^{2}

(x-1)

^{2}+(y+2)

^{2}=r

^{2}

x

^{2}-x-x+1+y

^{2}+2y+2y+4=r

^{2}

x

^{2}-2x+1+y

^{2}+4y+4=r

x

^{2}+y

^{2}-2x+4y+5=r

^{2}

I would really appreciate some direction on this.

Thank you all.

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