Which is the general form of the equation of the circle shown? x2 + y2 + 4x – 2y – 4 = 0 x2 + y2 + 4x – 2y + 2 = 0 x2 + y² – 4x + 2y – 4 = 0

The equation of a circle with center at point (h, k) and radius of r units is given by
[tex] (x-h)^{2} +(y-k)^2=r^2[/tex]

From the diagram, the center of the given circle is at points (-2, 1) and the radius is 3 units.

Therefore, the equation of the given by
[tex]\left(x-(-2)\right)^2+(y-1)^2=3^2 \\ \\ (x+2)^2+(y-1)^2=9 \\ \\ x^2+4x+4+y^2-2y+1=9 \\ \\ x^2+y^2+4x-2y+5-9=0 \\ \\ x^2+y^2+4x-2y-4=0[/tex]

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alokdubeyvidyaatech alokdubeyvidyaatech · Answer 2 · 2022-02-21T14:47:43+01:00

The general form of the equation of the circle with center (-2, 1) and radius 3 is [tex]x^2 + y^2 + 4x-2y-4=0[/tex].

Option 1 is the correct representation of the equation of the circle.

What is a circle?

The circle can be defined as a curved line that is the same distance from the center all the way around.

The general form of the equation of a circle with center (h, k) and radius r is given as,

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

From the diagram, the center of the given circle is at points (-2, 1) and the radius is 3 units. Then the equation of the circle is,

[tex](x - (-2))^2 + (y-1)^2 = 3^2[/tex]

[tex](x + 2)^2 + y^2 - 2y +1 =9[/tex]

[tex]x^2 + 4x+4 +y^2 -2y -8 =0[/tex]

[tex]x^2 + y^2 + 4x-2y-4=0[/tex]

Hence we can conclude that the general form of the equation of the circle is shown in option 1.

To know more about the circle., follow the link given below.

https://brainly.com/question/11833983.