1. ## circle

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

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

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

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

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: $2x^2 + 2y^2 - 16x + 4y = 1$

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

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

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

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

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

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

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

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

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

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

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

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