Respuesta :
we have that
x²-6x+y²+4y=d
Group terms that contain the same variable
(x²-6x)+(y²+4y)=d
Complete the square twice. Remember to balance the equation by adding the same constants to each side
(x²-6x+9)+(y²+4y+4)=d+9+4
the center of a circle is (3,-2)
the radius r²=d+13---------> 6²=d+13---------> d=36-13------> d=23
the answer is
the Y-coordinate of the center of the circle is -2
d is equal to 23
x²-6x+y²+4y=d
Group terms that contain the same variable
(x²-6x)+(y²+4y)=d
Complete the square twice. Remember to balance the equation by adding the same constants to each side
(x²-6x+9)+(y²+4y+4)=d+9+4
Rewrite as perfect squares
(x-3)²+(y+2)²=d+13the center of a circle is (3,-2)
the radius r²=d+13---------> 6²=d+13---------> d=36-13------> d=23
the answer is
the Y-coordinate of the center of the circle is -2
d is equal to 23
Part 1:
 For this case we have the following equation:
 x ^ 2 + y ^ 2 - 6x + 4y = d
 By completing squares we have:
 x ^ 2 + y ^ 2 - 6x + 4y + (-6/2) ^ 2 + (4/2) ^ 2 = d + (-6/2) ^ 2 + (4/2) ^ 2
 Rewriting we have:
 x ^ 2 + y ^ 2 - 6x + 4y + (-3) ^ 2 + (2) ^ 2 = d + (-3) ^ 2 + (2) ^ 2
 x ^ 2 + y ^ 2 - 6x + 4y + 9 + 4 = d + 9 + 4
 (x ^ 2 - 6x + 9) + (y ^ 2 + 4y + 4) = d + 13
 (x-3) ^ 2 + (y + 2) ^ 2 = d + 13
 Answer:
 The y coordinate of the center of the circle is:
 y = -2
 Part 2:
 For the value of d we have:
 d + 13 = 6
 d = 6 - 13
 d = -7
 Answer:
 The value of d is:
 d = -7
 For this case we have the following equation:
 x ^ 2 + y ^ 2 - 6x + 4y = d
 By completing squares we have:
 x ^ 2 + y ^ 2 - 6x + 4y + (-6/2) ^ 2 + (4/2) ^ 2 = d + (-6/2) ^ 2 + (4/2) ^ 2
 Rewriting we have:
 x ^ 2 + y ^ 2 - 6x + 4y + (-3) ^ 2 + (2) ^ 2 = d + (-3) ^ 2 + (2) ^ 2
 x ^ 2 + y ^ 2 - 6x + 4y + 9 + 4 = d + 9 + 4
 (x ^ 2 - 6x + 9) + (y ^ 2 + 4y + 4) = d + 13
 (x-3) ^ 2 + (y + 2) ^ 2 = d + 13
 Answer:
 The y coordinate of the center of the circle is:
 y = -2
 Part 2:
 For the value of d we have:
 d + 13 = 6
 d = 6 - 13
 d = -7
 Answer:
 The value of d is:
 d = -7