Prove algebraically that the straight line with equation x=2y+5 is a tangent to the circle with equation x²+y²=5.

x  = 2y + 5    (1)

x^2 + y^2  = 5  (2)

Sub (1) into (2) to find the y intersection of these functions

(2y + 5)^2  + y^2  =  5      simplify

4y^2 + 20 y + 25 + y^2  = 5

5y^2 + 20y +20  = 0      divide through by 5

y^2 + 4y + 4  =  0      factor

(y + 2)^2  = 0        take the square root of both sides

y + 2   = 0

y  = -2

And x  =    2(-2) + 5   =  1

So....(1, -2)  is the tangent point  because it  is the only point that makes both equations true

1 = 2(-2) + 5   is true      and

(1)^2  + (-2)^2  = 5     is also true

Here's a graph :  https://www.desmos.com/calculator/ty1aww3bmp