Find the tangent equation of the circle x 2 + y 2 = 4 from point a (2,4)

Find the tangent equation of the circle x 2 + y 2 = 4 from point a (2,4)

Obviously, x = 2 is one of the tangents; let y-4 = K (X-2), that is, kx-y + 4-2k = 0, and the distance from the center of the circle (0, 0) to the tangent is equal to the radius, then | 4 − 2K | K2 + 1 = 2, K = 34, 3x − 4Y + 10 = 0, and the tangent equation of the circle is x = 2, or 3x-4y + 10 = 0

Given the circle x square + y square = 4, find the tangent equation of point (2,5)

Point (2,5) is outside the circle. If the slope does not exist, then the tangent equation is x = 2. In this case, the distance from the center of the circle to the tangent is | 0-2 | = 2 = radius R. ② let the tangent equation be Y-5 = K (X-2), that is, kx-y + 5-2k = 0, the distance from the center of the circle to the tangent is d = | 0-0 + 5-2k | / √ (K & # 178; + 1) = 2, then the solution of 25-20k + 4K & # 178; = 4K & # 178; + 4 is obtained

Given the equation of a circle and the coordinates of point P, the equation of a line tangent to point P is obtained
Find the simplest way

Undetermined coefficient method
The linear equation is set up in the form of point slant,
Use the distance from the center of the circle to the straight line = radius to get the equation, find out the slope and substitute it into the equation
Pay attention
(1) P has a solution on the circle
(2) P has two solutions outside the circle. If two solutions are obtained from the solution equation, that is, the slope of the straight line,
If a solution is obtained, the slope of another straight line does not exist, and the tangent angle is 90 degrees

A ray from point a (- 2,3) is reflected by x-axis and tangent to circle C: (x-3) ^ 2 + (Y-2) ^ 2 = 1. The equation of the straight line of the reflected ray is obtained

The symmetry point of a (- 2,3) about X axis is (- 2, - 3)
So the reflected light is a straight line passing through the point (- 2, - 3), let y + 3 = K (x + 2), that is, kx-y + 2k-3 = 0
Because it is tangent to a circle, so | 3K-2 + 2k-3 | / √ (K & # 178; + 1) = 1
The solution is k = 4 / 3 or K = 3 / 4
So the linear equation is 4x-3y-1 = 0 or x-4y-6 = 0