Prove that if the point m, –im, 1 are collinear , then m lies on a circle.
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To prove that if the points m, -im, and 1 are collinear, then m lies on a circle, we need to show that these points satisfy the equation of a circle.
A circle can be represented by the equation (x - h)^2 + (y - k)^2 = r^2, where (h, k) are the coordinates of the center of the circle, and r is the radius.
Let's assume that m lies on the circle with center (h, k) and radius r. Therefore, the coordinates of m can be written as (x, y) = (m, -im).
Substituting these values into the equation of the circle, we get:
(x - h)^2 + (y - k)^2 = r^2
Substituting (x, y) = (m, -im) and simplifying:
(m - h)^2 + (-im - k)^2 = r^2
Expanding the squares:
m^2 - 2mh + h^2 + i^2m^2 + 2imk + k^2 = r^2
Since -1 multiplied by itself is equal to 1:
m^2 - 2mh + h^2 + m^2 + 2imk + k^2 = r^2
Grouping the real and imaginary terms:
(2m^2 + h^2 + k^2) + (-2mh + 2imk) = r^2
Now, since the points m, -im, and 1 are collinear, their slopes will be equal. We can find the slope between the points m and -im as:
Slope = (-im - (-m))/(m - m) = (-im + im)/(0) = 0/0
Here, we have an indeterminate form of 0/0. To resolve this, we differentiate each term of the equation with respect to m.
d/dm [(2m^2 + h^2 + k^2) + (-2mh + 2imk)] = d/dm (r^2)
Simplifying the left side:
d/dm (2m^2 + h^2 + k^2) + d/dm (-2mh + 2imk) = 0
4m - 2h + 2ik = 0
Now, equating the real and imaginary parts to zero:
4m - 2h = 0 (Equation 1)
2ik = 0 (Equation 2)
From Equation 2 (2ik = 0), we get k = 0, since i is a non-zero constant.
Substituting k = 0 into Equation 1 (4m - 2h = 0), we get:
4m - 2h = 0
4m = 2h
m = h/2
Therefore, we have found that the slope between the points m and -im is independent of m, which means the points lie on a straight line. Now, substituting k = 0 and m = h/2 into the original equation of the circle:
(x - h)^2 + (y - k)^2 = r^2
(x - h)^2 + y^2 = r^2
Since y = -im, we can also write it as:
(x - h)^2 + (-im)^2 = r^2
(x - h)^2 + m^2 = r^2
This equation represents a circle centered at (h, 0) with radius r, which proves that if the points m, -im, and 1 are collinear, then m lies on a circle.