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

Question

Neporo4naja [7]

3 years ago

6

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

Mathematics

2 answers:

Nadusha1986 [10] · Answer 1 · 2022-04-29T17:39:17+03:00

Nadusha1986 [10]3 years ago

6 0

Answer:

a

Step-by-step explanation:

Lunna [17] · Answer 2 · 2022-04-29T15:51:51+03:00

Lunna [17]3 years ago

3 0

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

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
$\left(x-(-2)\right)^2+(y-1)^2=3^2 \\ \\ (x+2)^2+(y-1)^2=9 \\ \\ x^2+4x+4+y^2-2y+1=9 \\ \\ x^2+y^2+4x-2y+5-9=0 \\ \\ x^2+y^2+4x-2y-4=0$