-3 and 3, both three away from 0
We have the following equation:
x2 + y2 + 42x + 38y - 47 = 0
We rewrite the equation:
x2 + 42x + y2 + 38y - 47 = 0
x2 + 42x + y2 + 38y = 47
Rewriting we have:
x2 + 42x + (42/2) ^ 2 + y2 + 38y + (38/2) ^ 2 = 47 + (42/2) ^ 2 + (38/2) ^ 2
x2 + 42x + 441 + y2 + 38y + 361 = 47 + 441 + 361
Rewriting we have:
(x + 21) ^ 2 + (y + 19) ^ 2 = 849
The center of the circle is:
(x, y) = (-21, -19)
The radio is:
r = root (849)
r = (849) ^ 2
A circle of the same radius is given by:
x ^ 2 + y ^ 2 - 50x - 30y + 1 = 0
Let's check:
x ^ 2 - 50x + y ^ 2 - 30y + 1 = 0
x ^ 2 - 50x + y ^ 2 - 30y = - 1
x ^ 2 - 50x + (-50/2) ^ 2 + y ^ 2 - 30y + (-30/2) ^ 2 = - 1 + (-30/2) ^ 2 + (-50/2) ^ 2
x ^ 2 - 50x + (-50/2) ^ 2 + y ^ 2 - 30y + (-30/2) ^ 2 = - 1 + 225 + 625
(x-25) ^ 2 + (y-15) ^ 2 = 849
Answer:
(x + 21) ^ 2 + (y + 19) ^ 2 = 849
(x, y) = (-21, -19)
r = (849) ^ 2
x ^ 2 + y ^ 2 - 50x - 30y + 1 = 0
3.2E-9
A bit complicated, so PLEASE don't ask for an explanation
the answer to your question is -12
The form of quadratic representing the roots can be expressed as,
Where p and q are the roots of the equation.
The form which represents the zeroes of function h is,
Where 2 and -2 are the roots of the function h(x).
Thus, option (B) is correct.
