You need to solve the system of equations, such that:
`{(x+y-5=0),(x^2+y^2-2x-4y+3=0):} =>
`{(x = 5 - y),((5 - y)^2 + y^2 - 2(5 - y) - 4y + 3 = 0):}` `
Expanding the square in the second equation yields:
`25 - 10y + y^2 - 10 + 2y -...
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Already a member? Log in here.
You need to solve the system of equations, such that:
`{(x+y-5=0),(x^2+y^2-2x-4y+3=0):} =>
`{(x = 5 - y),((5 - y)^2 + y^2 - 2(5 - y) - 4y + 3 = 0):}` `
Expanding the square in the second equation yields:
`25 - 10y + y^2 - 10 + 2y - 4y + 3 = 0`
`y^2 - 12y + 18 = 0`
You should use quadratic formula, such that:
`y_(1,2) = (12 +- sqrt(12^2 - 4*1*18))/2`
`y_(1,2) = (12 +- sqrt(72))/2`
`y_(1,2) = (12 +- 6sqrt(2))/2 => y_(1,2) = 6 +- 3sqrt2`
`x_(1,2) = 5 - y_(1,2) => x_(1,2) = 5 - (6 +- 3sqrt2)`
`x_1 = -1 - 3sqrt2`
`x_2 = -1 + 3sqrt2`
Hence, since the line `x + y - 5 = 0 ` intersects the circle in two points, `(-1-3sqrt2,6+3sqrt2), (-1+3sqrt2,6-3sqrt2)` , the given line is not tangent to the circle, but secant.