The equation of the tangent at a given point on a circle can be determined using the concept of the derivative. The derivative represents the slope of the tangent line at any given point on a curve.

Given circle equation: x^2 + y^2 - 4x - 2y - 20 = 0

To find the equation of the tangent at a specific point, we need to find the derivative of the circle equation and substitute the coordinates of the given point into the equation of the tangent.

Step 1: Find the derivative of the circle equation
To do this, we differentiate each term with respect to x, treating y as a constant.

Differentiating x^2 with respect to x gives us 2x,
Differentiating y^2 with respect to x gives us 2yy'.

The terms -4x and -2y do not involve x explicitly, so their derivative will be 0.

Therefore, the derivative of the circle equation with respect to x is:
2x + 2yy' - 4 = 0

Step 2: Substitute the coordinates of the given point
Given point: P(10, 7)

Substituting x = 10 and y = 7 into the derivative equation:
2(10) + 2(7)y' - 4 = 0
20 + 14y' - 4 = 0
14y' = -16
y' = -16/14
y' = -8/7

Step 3: Determine the equation of the tangent
The equation of a line can be represented in the slope-intercept form: y = mx + c, where m is the slope of the line.

We have the slope, y' = -8/7, and the point P(10, 7).
Substituting these values into the slope-intercept form, we get:
y = (-8/7)x + c

To find the value of c, substitute the coordinates of point P(10, 7):
7 = (-8/7)(10) + c
7 = -80/7 + c
7 + 80/7 = c
(49 + 80)/7 = c
129/7 = c

Therefore, the equation of the tangent at the point P(10, 7) is:
y = (-8/7)x + 129/7

Simplifying this equation, we can multiply through by 7 to eliminate the fraction:
7y = -8x + 129

Finally, rearranging the equation to the standard form, we get:
8x + 7y - 129 = 0

So, the equation of the tangent at the nearest point from P(10, 7) on the circle x^2 + y^2 - 4x - 2y - 20 = 0 is 8x + 7y - 129 = 0.