Find the equ. of a code passing through the (5, -2) and has the same centre as the circle x ^ 2 + y ^ 2 - 4x + 8y - 40 = 0
First, we need to find the center of the circle given by the equation x^2 + y^2 - 4x + 8y - 40 = 0. We can rewrite this equation in standard form by completing the square for x and y terms:
(x^2 - 4x) + (y^2 + 8y) = 40
(x^2 - 4x + 4) + (y^2 + 8y + 16) = 40 + 4 + 16
(x - 2)^2 + (y + 4)^2 = 60
From this equation, we can see that the center of the circle is at (2, -4) and the radius is √60.
Since the line passes through the point (5, -2) and has the same center as the circle, it will have the same slope as the diameter of the circle which connects the center point (2, -4) and the point (5, -2).
The slope of the diameter can be found by the formula (y2-y1)/(x2-x1), where (x1, y1) = (2, -4) and (x2, y2) = (5, -2):
m = (-2 - (-4)) / (5 - 2)
m = 2 / 3
So, the equation of the line passing through the points (5, -2) with a slope of 2/3 is given by the point-slope form:
y - y1 = m(x - x1)
y + 2 = (2/3)(x - 5)
y + 2 = (2/3)x - 10/3
y = (2/3)x - 16/3
Therefore, the equation of the line passing through the point (5, -2) and having the same center as the circle x^2 + y^2 - 4x + 8y - 40 = 0 is y = (2/3)x - 16/3.