Find the center and radius of circle

**zorro** · December 28th 2009, 01:11 PM

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?

**bigwave** · December 28th 2009, 01:32 PM

did you change this expression to to

$<br /> (x-h)^2 + (y-k)^2 = r^2<br />$

center is $(h,k)$ and radius $r>0$

**zorro** · December 28th 2009, 02:08 PM

No !!!!!

just used the general equation and found the center and raduis through it?????

**Prove It** · December 28th 2009, 03:45 PM

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 $y$ .

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

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

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

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

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

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

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

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

Can you read off the centre and the radius now?

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