Question

1. The equation of a circle is (x + 3)2 + (y + 5)2 = 8. Determine the coordinates of the center of the circle.

Answer 1

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 2

Answer:

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

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

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

4)

Standard equation of circle  $(x+1)^{2} +(y-2)^{2}$=$1^{2}$.

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 $(x+3)^{2} +(y+5)^{2}$=8.

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

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 $(x-3)^{2} +(y-2)^{2}$=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  $(x+1)^{2} +(y-2)^{2}$=$1^{2}$.