Identify the center and radius of the circle with the following equation: [tex]x^2+y^2+4x+8y+11=0[/tex]

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:

[tex] \displaystyle \text{centre} :( - 2, - 4)[/tex]

[tex] \text{redius} : 3[/tex]

Step-by-step explanation:

we are given a circle equation

[tex] \rm \displaystyle {x}^{2} + {y}^{2} + 4x + 8y + 11 = 0[/tex]

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

[tex] \rm \displaystyle {x}^{2} + {y}^{2} +A x +B y +C = 0[/tex]

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

[tex] \displaystyle (x - h {)}^{2} + (x - k {)}^{2} = {r}^{2} [/tex]

where:

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:

[tex] \rm \displaystyle ( {x}^{2} + 4x ) +( {y}^{2} + 8y) + 11 = 0[/tex]

cancel 11 from both sides:

[tex] \rm \displaystyle ( {x}^{2} + 4x ) +( {y}^{2} + 8y) = - 11[/tex]

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

[tex] \rm \displaystyle ( {x}^{2} + 4x + 4) +( {y}^{2} + 8y) = -11 + 4[/tex]

simplify addition:

[tex] \rm \displaystyle ( {x}^{2} + 4x + 4) +( {y}^{2} + 8y) = - 7[/tex]

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

[tex] \rm \displaystyle ( {x}^{2} + 4x + 4) +( {y}^{2} + 8y + 16) = 9[/tex]

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

[tex] \rm \displaystyle ( {x}+ 2) ^{2} +( {y} + 4) ^{2} = 9[/tex]

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

[tex] \rm \displaystyle ( {x} - ( - 2)) ^{2} +( {y} -( - 4) ) ^{2} = {3}^{2} [/tex]

therefore we obtain:

and we are done!

Please help! In the diagram below, lines l and k are parallel.




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