Question

question 6 of 10
what is the radius of a circle with the equation $x^{2}+y^{2}+2x + 4y-9 = 0$?
round your answer to the nearest thousandth.

Explanation:

Step1: Rewrite the equation in standard form

The standard form of a circle equation is $(x - a)^2+(y - b)^2=r^2$. Complete the square for $x$ and $y$ terms in $x^{2}+y^{2}+2x + 4y-9 = 0$.
For the $x$ - terms: $x^{2}+2x=(x + 1)^{2}-1$.
For the $y$ - terms: $y^{2}+4y=(y + 2)^{2}-4$.
So the equation becomes $(x + 1)^{2}-1+(y + 2)^{2}-4-9=0$.

Step2: Simplify the equation

$(x + 1)^{2}+(y + 2)^{2}-14 = 0$.
Then $(x + 1)^{2}+(y + 2)^{2}=14$.

Step3: Identify the radius

Comparing with the standard form $(x - a)^2+(y - b)^2=r^2$, we have $r^{2}=14$. So $r=\sqrt{14}\approx3.742$.

Answer:

$3.742$