Answer:
Step-by-step explanation:
You want the equations of tangents to x² +y² = 4 where the slope is 1.
The slope at a point (x, y) on the circle will be ...
2x·dx +2y·dy = 0
dy/dx = -2x/(2y) = -x/y
For the slope to be 1, the tangent points will lie on the line ...
-x/y = 1
y = -x
The values of x at those points will be ...
x² +(-x)² = 4 . . . . . . . . . . substitute for y
x² = 2 . . . . . . . . . . . . . divide by 2
x = {-√2, +√2} . . . . take the square roots
y = -x = {√2, -√2}
Using the point-slope equation for a line, we find the equations for the tangent lines to be ...
y -k = m(x -h) . . . . . . line with slope m through point (h, k)
y -√2 = 1(x -(-√2)) ⇒ x - y = -2√2
and
y -(-√2) = 1(x -√2) ⇒ x -y = 2√2
The equations for the tangent lines are x - y = ±2√2.