## anonymous one year ago Find the center (h, k) and the radius r of the circle 2 x^2 - 6 x +2 y^2 + 9 y - 3 = 0 .

1. zzr0ck3r

You need to complete the square twice. I will do one of them, $2 x^2 - 6 x +2 y^2 + 9 y - 3 = 0\\2x^2-6x+2y^2+9y=3\\2(x^2+3x)+2y^2+9y=3\\2(x+\frac{3}{2})^2+2y^2+9y=3+2(\frac{3}{2})^2$

2. anonymous

why is it 3/2?

3. zzr0ck3r

Equation of a circle is $(x-h)^2+(y-k)^2=r^2$ Where $$(h,k)$$ is the center and $$r$$ is the radius. Yuu have to complete the square twice. $2 x^2 - 6 x +2 y^2 + 9 y - 3 = 0 \\2x^2-6x+2y^2+9y=3 \\2(x^2-3x)+2(y^2+\frac{9}{2}y)=3 \\2(x-\frac{3}{2})^2+2(y+\frac{9}{4})^2=3+2(\frac{3}{2})^2+2(\frac{9}{4})^2=\frac{141}{8}\\\$$x-\frac{3}{2})^2+(y+\frac{9}{4})^2=\frac{141}{16}$ 4. anonymous ohhh okay that makes a lot of sense.. 5. anonymous wait, ive tried it again but for radius i got 117/8. I tried to enter both but none of the radius were right @zzr0ck3r 6. zzr0ck3r The radius would be, according to the formula, \(r=\sqrt{\frac{141}{16}}=\frac{\sqrt{141}}{4}$$.