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