Question

Identify the center and radius of the circle with the following equation: B8y%2B11%3D0" id="TexFormula1" title="x^2+y^2+4x+8y+11=0" alt="x^2+y^2+4x+8y+11=0" align="absmiddle" class="latex-formula">

Answer 1

Hey there,

Put the equation into the circle equation [ (x - h)² + (y - k)² = r² ].

x² + y² + 4x + 8y = -11

(x² + 4x) + (y² + 8y) = -11

(x + 2)² + (y + 4)² = 9

If that is confusing, you can always rewrite it to show exactly what you need in the circle equation.

(x - (-2))² + (y - (-4))² = 3²

Therefore, using the circle equation we can see that the center is at the point (-2, -4) and the radius is 3.

Best of Luck!

Answer 2

Answer:

$\displaystyle \text{centre} :( - 2, - 4)$

$\text{redius} : 3$

Step-by-step explanation:

we are given a circle equation

$\rm \displaystyle {x}^{2} + {y}^{2} + 4x + 8y + 11 = 0$

we want to figure out the centre and the redious of the circle from the equation notice that the given circle equation is in general form given by

$\rm \displaystyle {x}^{2} + {y}^{2} +A x +B y +C = 0$

general form of circle equation is a simplified version of standard form. the standard form of circle equation is given by:

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

where:

• (h,k) is the centre
• r is the redious

notice that the there're two completed squares in the standard form therefore we can consider completing square to figure out the centre and the redious

indentifying centre and redious:

rearrange:

$\rm \displaystyle ( {x}^{2} + 4x ) +( {y}^{2} + 8y) + 11 = 0$

cancel 11 from both sides:

$\rm \displaystyle ( {x}^{2} + 4x ) +( {y}^{2} + 8y) = - 11$

add 4 to the both sides so that the equality don't affect:

$\rm \displaystyle ( {x}^{2} + 4x + 4) +( {y}^{2} + 8y) = -11 + 4$

simplify addition:

$\rm \displaystyle ( {x}^{2} + 4x + 4) +( {y}^{2} + 8y) = - 7$

once again to complete the square add 16 to both sides which yields:

$\rm \displaystyle ( {x}^{2} + 4x + 4) +( {y}^{2} + 8y + 16) = 9$

complete the squares by using (a+b)²=a²+2ab+b² pattern:

$\rm \displaystyle ( {x}+ 2) ^{2} +( {y} + 4) ^{2} = 9$

we can still rewrite it to figure out the redious and the centre:

$\rm \displaystyle ( {x} - ( - 2)) ^{2} +( {y} -( - 4) ) ^{2} = {3}^{2}$

therefore we obtain:

• centre:(h,k)=(-2,-4)
• redious:r=3

and we are done!