<h2>
Answer and step by step explanation</h2>
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
