Question

5-4-3-2
16
6/ +
77?1?
x
Which is the general form of the equation of the circle
shown?
Ox²+²+4x-2y-4 = 0
Ox+y+4x-2y + 2 = 0
Ox² + y² 4x +2y-4 = 0
Ox² + y² 4x + 2y + 2 = 0

Answer 1

Answer:

1st option

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r is the radius

here (h, k ) = (- 2, 1 ) and r = 3 , then

(x - (- 2) )² + (y - 1)² = 3² , that is

(x + 2)² + (y - 1)² = 9 ← expand factors using FOIL

x² + 4x + 4 + y² - 2y + 1 = 9

x² + 4x + y² - 2y + 5 = 9 ( subtract 9 from both sides )

x² + 4x + y² - 2y - 4 = 0 , that is

x² + y² + 4x - 2y - 4 = 0 ← in general form

Answer 2

Answer:

$\textsf{1)} \quad x^2+y^2+4x-2y-4=0$

Step-by-step explanation:

Equation of a circle

$(x-a)^2+(y-b)^2=r^2$

where:

• (a, b) is the center
• r is the radius

From inspection of the graph:

• center of the circle = (-2, 1)
• radius of the circle = 3

Substitute the found values into the formula:

$\implies (x-(-2)^2+(y-1)^2=3^2$

$\implies (x+2)^2+(y-1)^2=9$

Expand and simplify:

$\implies (x+2)^2+(y-1)^2=9$

$\implies x^2+4x+4+y^2-2y+1=9$

$\implies x^2+y^2+4x-2y+5=9$

$\implies x^2+y^2+4x-2y-4=0$

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