Find the coordinates of the point(s) of intersection of the line y = 2x + 5 with the circle (x - 3)^2 + (y + 4)^2 = 50.

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- Jun 16th 2009, 01:39 AMMr RayonFind the coordinates...
Find the coordinates of the point(s) of intersection of the line y = 2x + 5 with the circle (x - 3)^2 + (y + 4)^2 = 50.

- Jun 16th 2009, 02:04 AMflyingsquirrel
Where are you stuck ?

- Jun 16th 2009, 02:07 AMMr Rayon
- Jun 16th 2009, 02:14 AMyeongil
Well you have two equations and two unknowns. Start by plugging in y = 2x + 5 into the other equation, like this:

$\displaystyle (x - 3)^2 + ({\color{red}2x + 5} + 4)^2 = 50$.

Solve for x, and when you're done, plug that back into y = 2x + 5 to solve for y. (You may get two answers for x, so plug each answer in to get a different y.)

01 - Jun 16th 2009, 02:31 AMMr Rayon
(x - 3)^2 + (2x + 5 + 4)^2 = 50

(x - 3)^2 + (2x + 9)^2 = 50

x^2 - 6x + 9 + 4x^2 + 36x + 81 = 50

5x^2 + 30x + 90 = 50

5x^2 + 30x + 40 = 0

Using the quadratic formula:

x = -2 or x = -4

y = 2 (-2) + 5

= 1

or

y = 2(-4) + 5

= -3

Am I right so far? If so, what do I do next? - Jun 16th 2009, 03:06 AMyeongil
That's it. You have the points: (-2, 1) and (-4, -3).

BTW, you didn't have to use the quadratic formula -- the quadratic was factorable. (Wink)

$\displaystyle \begin{aligned}

5x^2 + 30x + 40 &= 0 \\

5(x^2 + 6x + 8) &= 0 \\

5(x + 2)(x + 4) &= 0 \\

...

\end{aligned}$

01