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1. The equation of a circle is (x + 3)2 + (y + 5)2 = 8. Determine the coordinates of the center of the circle.
2. Write the standard equation of the circle below.
(First Image)
3. Using the circle below, determine the coordinates of the center and the length of the radius.
(Second Image)
4. Write the standard equation of the circle below.
(Third Image)
Thank you for the help. Happy holidays

1 The equation of a circle is x 32 y 52 8 Determine the coordinates of the center of the circle 2 Write the standard equation of the circle below First Image 3 class=
1 The equation of a circle is x 32 y 52 8 Determine the coordinates of the center of the circle 2 Write the standard equation of the circle below First Image 3 class=
1 The equation of a circle is x 32 y 52 8 Determine the coordinates of the center of the circle 2 Write the standard equation of the circle below First Image 3 class=

Respuesta :

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1. (-3, -5)
2. (x-3)2+(y-2)2=4
3. (-2,1), r=3
4. (x+1)2+(y-2)2=1
To find an equation for a circle from its graph, just look at the circle and if you have a centre at a point P(a,b) and a radius r, then your equation will be (x-a)2+(y-b)2=r2. Use the reverse process to draw a graph of a circle from its equation.

Answer:

1) Coordinates of center of circle of given equation is (-3,-5).

2) The equation of a circle is [tex](x-3)^{2} +(y-2)^{2}[/tex]=4.

3)Coordinates of center of circle of graph 2 are( -2,1) , radius of circle = 3.

4)

Standard equation of circle  [tex](x+1)^{2} +(y-2)^{2}[/tex]=[tex]1^{2}[/tex].

Step-by-step explanation:

Given: The equation of a circle is (x + 3)2 + (y + 5)2 = 8 and graph.

To find : Coordinates of center of circle and standard equation of of given circle .

Solution : We have [tex](x+3)^{2} +(y+5)^{2}[/tex]=8.

On comparing with the standard equation of circle  [tex](x-h)^{2} +(y-k)^{2}[/tex]=[tex]r^{2}[/tex].

Coordinates of center of circle are h,k and radius of circle =r .

1) Coordinates of center of circle of given equation is (-3,-5).

2) From graph 1 we can see coordinates of center of circle are (3,2)

The equation of a circle is [tex](x-3)^{2} +(y-2)^{2}[/tex]=4.

3)Coordinates of center of circle of graph 2 are( -2,1) and radius of circle = 3.

4) Coordinates of center of circle of graph 3 are( -1,2) and radius of circle = 1

then standard equation of circle  [tex](x+1)^{2} +(y-2)^{2}[/tex]=[tex]1^{2}[/tex].

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