It is known that the equation of a circle is (x-1) 2 + y2 = 4. A point a (4,2) (1) outside the circle solves the tangent equation of the circle passing through point a

It is known that the equation of a circle is (x-1) 2 + y2 = 4. A point a (4,2) (1) outside the circle solves the tangent equation of the circle passing through point a

Let the linear equation of a (4,2) be Y-2 = K (x-4), that is, kx-y + 2-4k = 0
The distance from the center of the circle to the straight line = radius, that is, k-0 + 2-4k | / radical (k ^ 2 + 1) = 2
|2-3K | = 2 radical (k ^ 2 + 1)
Square: 4-12k + 9K ^ 2 = 4K ^ 2 + 4
5k^2-12k=0
k(5k-12)=0
k=0,k=12/5
That is, the tangent equation is y = 2 and 12 / 5 x-y-38 / 5 = 0, that is, 12x-5y-38 = 0