# Thread: equation of circle

1. ## equation of circle

the question i have is: the points A(2, 5) and B(-4, 13) lie at opposite ends of a diameter of a circle with centre C

a) find the radius of the circle- i did this by finding the diameter which was 10 and halving it, so the radius is 5

b) (which is the one i had a problem with): find the equation of the circle in the form $x^2 + y^2 + px + qy + c = 0$

i found that the centre was (-1, 9) like so $\frac{2 + (-4)}{2} = -1$ and $\frac {5 + 13}{2} = 9$

so the equation of the circle would be this: $(x + 1)^2 + (y - 9)^2 = 25$

the answer the book gave was $x^2 + y^2 + 2x - 18y + 57 = 0$ which would imply that you made the squared constant terms positive, but it thought you would make them negative ie $(x + 1)^2 - 1 + (y - 9)^2 - 81 = 25$

does the method i used only work for certain things?

thanks

2. Originally Posted by mark
the question i have is: the points A(2, 5) and B(-4, 13) lie at opposite ends of a diameter of a circle with centre C

a) find the radius of the circle- i did this by finding the diameter which was 10 and halving it, so the radius is 5

b) (which is the one i had a problem with): find the equation of the circle in the form $x^2 + y^2 + px + qy + c = 0$

i found that the centre was (-1, 9) like so $\frac{2 + (-4)}{2} = -1$ and $\frac {5 + 13}{2} = 9$

so the equation of the circle would be this: $(x + 1)^2 + (y - 9)^2 = 25$

the answer the book gave was $x^2 + y^2 + 2x - 18y + 57 = 0$ which would imply that you made the squared constant terms positive, but it thought you would make them negative ie $(x + 1)^2 - 1 + (y - 9)^2 - 81 = 25$

does the method i used only work for certain things?

thanks

Just open up parentheses in the circle's equation!

$(x+1)^2+(y-9)^2=81\Longrightarrow x^2+2x+y^2-18y+82=25...$

Tonio