Find both points where the line y=2x-1 intersects a circle centered at the origin with radius 3. (round to the three decimal places)

Answer 1

The equation for a circle with center at the origin with radius 3 is x² + y² = 3² or x² + y² = 9.

The intersection point would be a point that complies with both y = 2x - 1 AND x² + y² = 9 meaning that the x and the y are the same for both. So, since they are the same, we can substitute "2x - 1" into x² + y² = 9 wherever there is a "y" so:

x² + (2x - 1)² = 9

x² + (4x² - 4x + 1) = 9

5x² - 4x - 8 = 0

Using the quadratic formula, for ax² + bx + c, x = [-b ± √(b² - 4ac)]/(2a)

a = 5, b = -4, c = -8 so:

x = [-(-4) ± √((-4)² - 4)5)(-8))]/(2(5)) = [4 ± √16+160)]/10 = (4 ± √176)/10 = (4 ± 13.266)/10

so x = either (4 + 13.266)/10 or (4 - 13.266)/10 so x = 1,727 or x = -0.927

To find the points, we also need to y-values. To find those we have y = 2x - 1 so when x = 1,727, y = 2(1,727) - 1 or y = 2.453 making the ordered pair (1,727, 2.453). When x = -0.927, y = 2(-0.927) - 1 or y = -2.853 making the ordered pair (-0.927, -2.853)