circle problem?

Question

equation: x^2 + y^2 +4x +2y - 20 =0

a) find the center

b)find the radius

c) find the intercepts


please help me, I have a few problems like this, thanks!

Answer 1

x^2 + y^2 + 4x + 2y - 20 = 0

to make life easy you need to convert the equation into standard form

(x^2 + 4x + 4) + (y^2 + 2y + 1) - 20 - 5 = 0
(x + 2)^2 + (y + 1)^2 = 25

Now we can easily see that the center is at (-2, -1) and the radius is 5 (an earlier answerer showed you how THAT works).

Answer 2

There are standard formulas for centre and radius, but I'll explain the method:
The idea is to get the equation in the form
(x - ...)^2 + (y - ...)^2 = positive constant,
and then the values missing above are
x coord of centre, y coord of centre, radius^2

First write the equation as
x^2 + 4x + y^2 + 2y = 20
To "complete the square" on x^2 + 4x, calculate
(4/2)^2, which is 4, so we add 4 on the left and also on the right to keep it balanced.
For y^2 + 2y, find (2/2)^2 which is 1, so we add 1 to both sides. Thus the equation is
x^2 + 4x + 4 + y^2 + 2y + 1 = 20 + 4 + 1
i.e.
(x + 2)^2 + (y + 1)^2 = 25

Hence the coordinates of the centre are (-2, -1),
and the radius is √25 = 5

c) To find the intercepts on the x and y axes:

On the x axis, y = 0, so sub y = 0 and solve for x:
x^2 + 4x - 20 = 0
This equation hasn't rational solutions, so use the formula:
x = [-4 ± √(4^2 - 4*1*(-20))]/(2*1)
or, preferably, the central form we've found: i.e. sub y = 0 in
(x + 2)^2 + (y + 1)^2 = 25
and get
(x + 2)^2 + 1^2 = 25
(x + 2)^2 = 24
x + 2 = ± √24 = ± 2√6
Hence x = -2 + 2√6 or -2 - 2√6
[i.e. about2.9, and about -6.9]

Likewise, for the intercepts on the y axis, sub x = 0 and get
2^2 + (y + 1)^2 = 25
and so
(y + 1)^2 = 21
Hence y = -1 ± √21
[about 3.6, and about -5.6]

Thus you have the four points where this circle cuts the x and y axes.

Answer 3

Equation of circle is:-
x² + y² + 2gx + 2fy + c = 0
x² + y² + 4x + 2y - 20 = 0 is given circle
2g = 4
g = 2
2f = 2
f = 1
c = - 20
a) Centre (-g,-f) is (- 2,-1)

b) r = √(g² + f² - c) = √(4 + 1 + 20) = 5

c) Cuts x axis at:-
x² + 4x - 20 = 0
x = [- 4 ± √96 ] / 2
x = [- 4 ± 4√6 ] / 2
x = [- 2 ± 2√6 ]
x = 2.[- 1 ± √6]
Cuts y axis at:-
y² + 2y - 20 = 0
y = [- 2 ± √84]/2
y = [- 2 ± 2√21 ] / 2
y = - 1 ± √21

Answer 4

you need to set it in the form

(x-k)^2 + (y-l)^2 = r^2

where r is the radius, and (k,l) is the center, so

x^2 + y^2 +4x + 2y - 20 = 0
x^2 + 4x + y^2 + 2y = 20
x^2 + 4x + 4 + y^2 + 2y + 1 = 20 + 4+ 1
(x+2)^2 + (y+1)^2 = 25

so the center is at (-2,-1)
radius = 5

for y intercept, let x = 0, so

(0+2)^2 + (y+1)^2 = 25
(y+1)^2 = 21
y+1 = +/- sqrt(21)
y = +/- sqrt(21) - 1

for x intercept, let y = 0, so

(x+2)^2 + (0+1)^2 = 25
(x+2)^2 = 24
x = +/-root (24) - 2

so intercepts are

( -root (24) , - 2 sqrt(21) - 1 )
( root (24) , - 2 sqrt(21) - 1 )
( -root (24) , 2 sqrt(21) - 1 )
( root (24) , 2 sqrt(21) - 1 )

Answer 5

x^2 + y^2 + 2gx +2fx +c =0

centre(-g,-f)
radius =√[(g)^2 +(f)^2 -c]

x^2 + y^2 +4x +2y - 20 =0
x^2 + y^2 +2(2)x + 2(1)y +(-20) =0

centre(-2 , -1)#

radius = √[2^2 +1^2 -(-20)]
=√[4+1+20]
=√25
=5#


when x=0,
0 + y^2 + 0 +2y -20 =0
y^2 +2y -20 =0
using formula
y = [-b+/-√(b² - 4ac)] / 2a
y = 3.583 or -5.583

when y=0,
x^2 +0 + 4x +0 -20 =0
x^2 +4x -20 =0
using formula
x = [-b+/-√(b² - 4ac)] / 2a
x=2.899 or -6.899

the intercepts,
(2.899, 0) , (-6.899 , 0)
(0 , 3.583) , (0 , -5.583)

Answer 6

a) ax^2+by^2+dx+ey+f=0 , c(h,k)
h=-d/2
k=-e/2
therefor h=-2 and k=-1 c(-2,-1)

b)r=1/2 squir root of d^2+e^2-4*f
1/2 *squir root of 4^2=2^2-4*-20
1/2 *squir root of 16+4+80
1/2 *squir root of 100
1/2 *10
5
Radius=5
c)you can find those with ease try to draw it on a graph

Answer 7

(x+2)^2+(y+1)^2=25
so thecenter is x=-2 y=-1
radius=5

didnt understand c


give me please best answerr