# Find the center and radius of circle

• December 28th 2009, 02:11 PM
zorro
Find the center and radius of circle
Question : Find the center and radius of the circle

$2x^2 + 2y^2 + 5x + 7y -3 = 0$

My Work:::::::::::

center $C(- \frac{5}{4} , - \frac{7}{2})$..............Is this correct?

Raduis = $\frac{7\sqrt{5}}{4}$......................Is this correct?
• December 28th 2009, 02:32 PM
bigwave
did you change this expression to to

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

center is $(h,k)$ and radius $r>0$
• December 28th 2009, 03:08 PM
zorro
No !!!!!

just used the general equation and found the center and raduis through it?????
• December 28th 2009, 04:45 PM
Prove It
Quote:

Originally Posted by zorro
Question : Find the center and radius of the circle

$2x^2 + 2y^2 + 5x + 7y -3 = 0$

My Work:::::::::::

center $C(- \frac{5}{4} , - \frac{7}{2})$..............Is this correct?

Raduis = $\frac{7\sqrt{5}}{4}$......................Is this correct?

You need to complete the square for $x$ and $y$.

$2x^2 + 2y^2 + 5x + 7y - 3 = 0$

$2x^2 + 5x + 2y^2 + 7y = 3$

$x^2 + \frac{5}{2}x + y^2 + \frac{7}{2}y = \frac{3}{2}$

$x^2 + \frac{5}{2}x + \left(\frac{5}{4}\right)^2 + y^2 + \frac{7}{2}y + \left(\frac{7}{4}\right)^2 = \frac{3}{2} + \left(\frac{5}{4}\right)^2 + \left(\frac{7}{4}\right)^2$

$\left(x + \frac{5}{4}\right)^2 + \left(y + \frac{7}{4}\right)^2 = \frac{49}{8}$

$\left[x - \left(-\frac{5}{4}\right)\right]^2 + \left[y - \left(-\frac{7}{4}\right)\right]^2 = \left(\sqrt{\frac{49}{8}}\right)^2$

$\left[x - \left(-\frac{5}{4}\right)\right]^2 + \left[y - \left(-\frac{7}{4}\right)\right]^2 = \left(\frac{7}{2\sqrt{2}}\right)^2$

$\left[x - \left(-\frac{5}{4}\right)\right]^2 + \left[y - \left(-\frac{7}{4}\right)\right]^2 = \left(\frac{7\sqrt{2}}{4}\right)^2$.