# Let x,y be real numbers and y^2+4y+9x^2-30x+29=0, which of the following is equal to 9x-y? A. 17 B. 25 C. 30 D. 41

Jun 17, 2018

A

#### Explanation:

You may notice that it bears some similarities to a circle with the general form ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$ where $\left(h , k\right)$ is the centre and r is the radius

So first up, you need to complete the square
${y}^{2} + 4 y + 9 {x}^{2} - 30 x + 29 = 0$

$\left(9 {x}^{2} - 30 x\right) + \left({y}^{2} + 4 y\right) = - 29$

$9 \left({x}^{2} - \frac{30}{9} x + {\left(\frac{5}{3}\right)}^{2}\right) + \left({y}^{2} + 4 y + 4\right) = - 29 + 4 + 25$
In case you don't remember how to complete the square,
$a {x}^{2} + b x + {\left(\frac{b}{2}\right)}^{2}$ is how you go about it. All you have to do to find your constant is to half the coefficient of your $x$ term ie $\frac{b}{2}$and then square the entire thing ie ${\left(\frac{b}{2}\right)}^{2}$

$9 {\left(x - \frac{5}{3}\right)}^{2} + {\left(y + 2\right)}^{2} = 0$

Therefore, the centre is $\left(\frac{5}{3} , - 2\right)$

Now you have the equation $9 x - y$. Sub your above point in and you will get:
$9 \left(\frac{5}{3}\right) - \left(- 2\right) = 15 + 2 = 17 = A$