# Thread: circle

1. ## circle

determine the coordinates of the centre and the radius of the circle with equation: $\displaystyle 2x^2 + 2y^2 - 16x + 4y = 1$

i started by making it $\displaystyle 2[x^2 - 8x + (-4)^2] + 2[y^2 + 2y + (1)^2] = 1 + (-8)^2 + (2)^2$ and then

$\displaystyle 2(x - 4)^2 + 2(y + 1)^2 = \sqrt69$ and i said the coordinates are $\displaystyle (8, -1)$ and the radius $\displaystyle \sqrt69$

the book says the radius is $\displaystyle \frac {1}{2} \sqrt70$ and the coordinates are $\displaystyle (4, -1)$

can someone tell me how i'm wrong please? thankyou

2. Originally Posted by mark
determine the coordinates of the centre and the radius of the circle with equation: $\displaystyle 2x^2 + 2y^2 - 16x + 4y = 1$

i started by making it $\displaystyle 2[x^2 - 8x + (-4)^2] + 2[y^2 + 2y + (1)^2] = 1 + (-8)^2 + (2)^2$ and then

$\displaystyle 2(x - 4)^2 + 2(y + 1)^2 = \sqrt69$ and i said the coordinates are $\displaystyle (8, -1)$ and the radius $\displaystyle \sqrt69$

the book says the radius is $\displaystyle \frac {1}{2} \sqrt70$ and the coordinates are $\displaystyle (4, -1)$

can someone tell me how i'm wrong please? thankyou
$\displaystyle 2x^2 + 2y^2 - 16x + 4y = 1$

$\displaystyle 2(x^2+y^2-8x+2y) = 1$

$\displaystyle (x^2+y^2-8x+2y) = \frac{1}{2}$

$\displaystyle (x-4)^2 - 16 + (y+1)^2-1 = \frac{1}{2}$

$\displaystyle (x-4)^2 + (y+1)^2 = \frac{33}{2}$

Centre is $\displaystyle (4,-1)$ and radius is $\displaystyle \sqrt{\frac{33}{2}}$

Not sure where $\displaystyle \frac{1}{2} \,\sqrt{70}$ comes from

3. isn't it $\displaystyle \sqrt\frac{35}{2}$?

4. Originally Posted by mark
isn't it $\displaystyle \sqrt\frac{35}{2}$?
A simple typo.