# 9y^2 + 7x^2 + 35 = 8x - 2x^2 - 36y find centre and radius?

• Apr 8th 2010, 06:27 PM
youmuggles
9y^2 + 7x^2 + 35 = 8x - 2x^2 - 36y find centre and radius?
rearrange this equation to show the centre-radius form of the equation of a circle. State the coordinates of the center and the radius of the circle

9y^2 + 7x^2 + 35 = 8x - 2x^2 - 36y

So far i have...

9y^2 + 7x^2 + 35 = 8x - 2x^2 - 36y

9x^2 - 8x + 9y^2 + 36y + 35 = 0

x^2 - (8/9)x + y^2 + 4y = -35/9

just a bit stuck now....
• Apr 8th 2010, 07:27 PM
Stroodle
Hi,

You need to complete the square for both $x$ and $y$:

$9x^2-8x+9y^2+36y+35=0$

$9(x^2-\frac{8}{9})+9(y^2+4y)+35=0$

$9\left [ (x-\frac{4}{9})^2-\frac{16}{81}\right ]+9\left [(y+2)^2-4\right ]+35=0$

$9(x-\frac{4}{9})^2-\frac{16}{9}+9(y+2)^2-36+35=0$

Can you take it from there?
• Apr 8th 2010, 07:43 PM
harish21
Quote:

Originally Posted by youmuggles
rearrange this equation to show the centre-radius form of the equation of a circle. State the coordinates of the center and the radius of the circle

9y^2 + 7x^2 + 35 = 8x - 2x^2 - 36y

So far i have...

9y^2 + 7x^2 + 35 = 8x - 2x^2 - 36y

9x^2 - 8x + 9y^2 + 36y + 35 = 0

x^2 - (8/9)x + y^2 + 4y = -35/9

just a bit stuck now....

$9y^2 + 7x^2 + 35 = 8x - 2x^2 - 36y$

$2x^2 -8x+ 9y^2 + 36y + 7x^2 + 35 = 0$

$9x^2 - 8x + 9y^2 +36y +36 -1 = 0$

$x^2 - \frac{8x}{9} + y^2 + 4y + 4 - \frac{1}{9} = 0$

$x^2 - \frac{8x}{9} + (y+2)^2= \frac{1}{9}$

$x^2 - \frac{8x}{9} +({\frac{4}{9}})^2 + (y+2)^2= \frac{1}{9}+ {{\frac{16}{81}}^2}$

$(x- {\frac{4}{9}})^2 + (y+2)^2 = \frac{25}{81}$

$(x- {\frac{4}{9}})^2 + (y+2)^2 = (\frac{5}{9})^2$

Compare the above equation with the equation of a circle, that is,

$(x-h)^2 + (y-k)^2 = r^2$

you have

center = (h,k) = $({\frac{4}{9}}, -2)$

radius = r = $\frac{5}{9}$