# Find the coordinates...

• Jun 16th 2009, 02:39 AM
Mr Rayon
Find 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, 03:04 AM
flyingsquirrel
Where are you stuck ?
• Jun 16th 2009, 03:07 AM
Mr Rayon
Quote:

Originally Posted by flyingsquirrel
Where are you stuck ?

From the beginning. I don't know what to do.

Do you think you can provide me a complete worked out solution to this?
• Jun 16th 2009, 03:14 AM
yeongil
Well you have two equations and two unknowns. Start by plugging in y = 2x + 5 into the other equation, like this:
$(x - 3)^2 + ({\color{red}2x + 5} + 4)^2 = 50$.

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, 03:31 AM
Mr 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, 04:06 AM
yeongil
Quote:

Originally Posted by Mr Rayon
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?

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)

\begin{aligned}
5x^2 + 30x + 40 &= 0 \\
5(x^2 + 6x + 8) &= 0 \\
5(x + 2)(x + 4) &= 0 \\
...
\end{aligned}

01