Question

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

Answer 1

x = 2y + 5 (1)

x^2 + y^2 = 5 (2)

Explanation:

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

Hope this helps :)