Question

Can you help me with these two questions please !!!!

Answer 1

Answer: $(x+2)^{2}+(y-4)^{2} =16$

and

$(x-6)^{2} +(y+4)^{2} =20$

Step-by-step explanation:

Equation for a Circle

$(x-h)^{2} +(y-k)^{2} =r^{2}$

h= -2 k= 4 r= 4

$(x+2)^{2}+(y-4)^{2} =16$

Part 2

Complete the Square

$x^{2} +y^{2} -12x+8y+32=0\\(x^{2}-12x+-)+(y^{2}+8y)=-32\\(x^{2} -12x+36)=(y^{2} +8y+16)=-32+36+16\\\\(x-6)^{2} +(y+4)^{2} =20$

Answer 2

Answer:

1) Choice B

(x-2)^2 + (y+4)^2 = 16

2) Choice A

(x-6)^2 + (y+4)^2 = 20

Step-by-step explanation:

Equation for circle is (x-h)^2 + (y-k)^2 = r^2

(h,k) is cords

1) (x-2)^2 + (y+4)^2 = 16

Cords = (-2, 4)

radius = $\sqrt{16}$ = 4

2) x^2 + y^2 - 12x +8y +32 = 0

x^2 - 12x + y^2 +8y +32 = 0

Choice A

(x-6)^2 + (y+4)^2 = 20

Multiply it out

x^2 - 12x + 36 + y^2 +8y + 16 = 20

Combine same values

x^2 - 12x + y^2 +8y + 52 = 20

Bring it all to one side (subtract)

x^2 - 12x + y^2 +8y +32 = 0