Question

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

Answer 1

You need to complete the square twice. I will do one of them,

\[2 x^2 - 6 x +2 y^2 + 9 y - 3 = 0\\2x^2-6x+2y^2+9y=3\\2(x^2+3x)+2y^2+9y=3\\2(x+\frac{3}{2})^2+2y^2+9y=3+2(\frac{3}{2})^2\]

Answer 2

why is it 3/2?

Answer 3

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.


\[2 x^2 - 6 x +2 y^2 + 9 y - 3 = 0
\\2x^2-6x+2y^2+9y=3
\\2(x^2-3x)+2(y^2+\frac{9}{2}y)=3
\\2(x-\frac{3}{2})^2+2(y+\frac{9}{4})^2=3+2(\frac{3}{2})^2+2(\frac{9}{4})^2=\frac{141}{8}\\\\(x-\frac{3}{2})^2+(y+\frac{9}{4})^2=\frac{141}{16}\]

Answer 4

ohhh okay that makes a lot of sense..

Answer 5

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

Answer 6

The radius would be, according to the formula, \(r=\sqrt{\frac{141}{16}}=\frac{\sqrt{141}}{4}\).