If line L y = 3 + x forms tangent to a circle with center at (1,-4), w...
We know that a tangent to a circle is perpendicular to the radius drawn to the point of tangency. Therefore, the line L, which is tangent to the circle, is perpendicular to the radius drawn from the center of the circle to the point of tangency.
The given equation of line L is y = 3 + x. We can rewrite it in the standard form as follows:
y - x = 3
The slope of line L is 1, as the coefficient of x is 1. The slope of a line perpendicular to L is the negative reciprocal of its slope. So the slope of the radius from the center of the circle to the point of tangency is -1.
We also know that the radius passes through the center of the circle, which is (1, -4). Using the point-slope form of a line, we can write the equation of the radius as:
y - (-4) = -1(x - 1)
y + 4 = -x + 1
y = -x - 3
Now, we need to find the equation of the other tangent to the circle that is parallel to line L. Since the tangent is parallel to line L, it will have the same slope of 1.
To find the equation of this parallel tangent line, we can use the point-slope form again, using the coordinates of the center of the circle (1, -4):
y - (-4) = 1(x - 1)
y + 4 = x - 1
y = x - 5
Comparing this equation with the given options, we find that the correct answer is:
(C) y = x - 13
Therefore, the equation of the other tangent to the circle that is parallel to line L is y = x - 13.