1: Find the linear equation with normal phasor (1, - 2) and tangent to circle x ^ 2 + y ^ 2-2y-4 = 0? 2: find the equation of circle with chord length of 2 √ 2 with P (2, - 1) as the center and cut by line x-y-1 = 0 1: The answer is x-2y + 7 = 0 or x-2y-3 = 0 2: The answer is (X-2) ^ 2 + (y + 1) ^ 2 = 4

1: Find the linear equation with normal phasor (1, - 2) and tangent to circle x ^ 2 + y ^ 2-2y-4 = 0? 2: find the equation of circle with chord length of 2 √ 2 with P (2, - 1) as the center and cut by line x-y-1 = 0 1: The answer is x-2y + 7 = 0 or x-2y-3 = 0 2: The answer is (X-2) ^ 2 + (y + 1) ^ 2 = 4

x^2+y^2-2y-4=0 ===>x^2+(y-1)^2=5
1. Let the tangent equation be y = K (x-1) - 2, that is, kx-y-k-2 = 0
The distance from the center of the circle to the straight line is r = | - 1-k-2 | / √ (k ^ 2 + 1) = √ 5
The solution is K1 = 2, K2 = - 1 / 2
So the linear equation is 2x-y-4 = 0 or x + 2Y + 3 = 0
2. Let the equation of circle (X-2) ^ 2 + (y + 1) ^ 2 = R ^ 2
The distance from the center of a circle to a straight line d = | 2 + 1-1 | / √ 2 = √ 2
The solution of R ^ 2 = D ^ 2 + (√ 2) ^ 2 is R ^ 2 = 4
So the circular equation is (X-2) ^ 2 + (y + 1) ^ 2 = 4
Your first answer is wrong!