2. Given the equation of a circle in standard form, identify the center, radius, and graph the circle.
(x + 1)2 + (y - 2)2 = 36

Center:
Radius:

2 Given the equation of a circle in standard form identify the center radius and graph the circle x 12 y 22 36 Center Radius class=

Respuesta :

Answer:

Center: ( -1 , 2 )

Radius: 6

Step-by-step explanation:

The equation for a circle is given as follow:

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

Where,

the Center is: ( h , k ) (note that the signs of the number are different)

and the radius is: r

So if we compare the original circle equation to the equation in the question we can see that:

[tex](x+1)^{2} +(y-2)^{2} =36[/tex]

the Center is: (-1,2)

and the radius is: [tex]\sqrt{36}[/tex] = 6

2. To draw the graph find points that lay on the circle, it's better to take the values of x and y from the Center:

first sub y=2 in the equation to find the values for x:

[tex](x+1)^{2} +(y-2)^{2} =36[/tex]

[tex](x+1)^{2} +(2-2)^{2} =36[/tex]

[tex](x+1)^{2} +(0)^{2} =36[/tex]

[tex](x+1)^{2} =36[/tex]

[tex]x+1 =±\sqrt{36}[/tex]

[tex]x=6-1[/tex]    AND    [tex]x=-6-1[/tex]

[tex]x=5[/tex]          AND    [tex]x=-7[/tex]

  • The points are A(5,2)  and  B(-7,2)

second sub x= -1 in the equation to find the values for y:

[tex](x+1)^{2} +(y-2)^{2} =36[/tex]

[tex](-1+1)^{2} +(y-2)^{2} =36[/tex]

[tex](0)^{2} +(y-2)^{2} =36[/tex]

[tex](y-2)^{2} =36[/tex]

[tex]y-2=±\sqrt{36}[/tex]

[tex]y=6+2[/tex]    AND    [tex]y=-6+2[/tex]

[tex]y=8[/tex]          AND    [tex]y=-4[/tex]

  • The points are D(-1,8)  and  E(-1,-4)      

After finding the points write them in the graph and match them together to get the like the circle in the picture below:

Ver imagen hajora03