If x square + y square of the circle is 25, and the tangent of the circle is taken at the last point m (- 3,4), then the tangent equation is?

If x square + y square of the circle is 25, and the tangent of the circle is taken at the last point m (- 3,4), then the tangent equation is?

Idea: according to the tangent, that is, the distance from the center of the circle to the straight line is d = r = 5, according to the formula of the distance from the point to the straight line, the steps are as follows: let the linear equation y = K (x-3) + 4 ① the center of the circle be (0,0) → d = | - 3K + 4 | / √ (1 + k) d = r = 5 → k = - 3 / 4  y = - 3 / 4 (x-3) + 4

Given the tangent equation of circle x square + y square = 25, passing through point a (4, - 3)
Forget, the main thing is the process

Point a is on the circle, a (4, - 3) is on the tangent, the two-point equation of the tangent is (y + 3) / (x-4) = (Y0 + 3) / (x0-4), ① the center of the circle is O, the OA vector is (4, - 3), the tangent vector is (x0-4, Y0 + 3), and OA is perpendicular to the tangent. Then 4 (x0-4) - 3 (Y0 + 3) = 0, that is, (Y0 + 3) / (x0-4) = 4 / 3, substituting into ①, we get (y + 3) / (x-4) = 4 / 3

Given x + y = 2, find the value of 3 (x + y) - 4x-4y + 5

3（x+y）-4x-4y+5
=3（x+y）-4（x+y）+5
=6-8+5
=3