1. ## Find the center and radius of circle

Question : Find the center and radius of the circle

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

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

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

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

2. did you change this expression to to

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

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

3. No !!!!!

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

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

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

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

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

Raduis = $\displaystyle \frac{7\sqrt{5}}{4}$......................Is this correct?
You need to complete the square for $\displaystyle x$ and $\displaystyle y$.

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

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

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

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

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

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

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

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