How to work out the Points of Intersection in Circle?

This is part of Core One for AS Level Maths and I totally understood it in the lesson; but now I'm home, I'm rather confused. xD

So if I have the equation of the line; x^2 - 8x + y^2 - 8y + 22 = 0

And the Circle Equation as; y= 2x + 1

Then we're told to find the co-ordinates of any points of intersection or tangential contact, or if the line fails to meet the curve - we must state this fact.

It should be straightforward right? I started off by substituting y= 2x + 1 into the Line Equation, like so:

x^2 - 8x + (2x + 1)- 8 (2x + 1) + 22 = 0

Then I expanded the brackets and simplified;

x^2 - 8x + 4x^2 + 4x + 2 - 16X + 8 + 22 = 0

5x^2 - 20x + 32 = 0

..Right? And it's about this point when I get stuck.

If we need to find out whether or not there are any points of intersection, we should be using the discriminant..? Which is b^2 - 4ac. Which I have done here;

b^2 - 4ac = -20^2 - 4 (5*32)

= 400 - 40*160

= 400 - 640

= -240

Which is less than 0; this means that there are no points of intersection?

I don't understand whether or not that is actually correct. There are four parts to this question; after just finishing part c it seems that every single one has no points of intersection. If this is the case; why would they ask me for the co-ordinates for the points of intersection?

Which leads me to believe I am doing these wrong - which might be what they want me to believe, but I figured I should try and double check my workings with someone.

So any comments would be greatly appreciated. :)

Posted Sat 6th October, 2012 @ 15:06 by Bexie

Firstly, the equation of the circle is x^2 - 8x + y^2 - 8y + 22, and the equation of the line is y=2x+1, not the other way around!

You have made a small mistake with the substitution in the fact that you forgot to square the y, hence it should be:

x^2 - 8x + (2x+1)^2 - 8(2x+1) + 22 = 0

which eventually expands to:

5x^2 - 20x + 15 = 0

Simplify this by dividing both sides by 5 to get:

x^2 - 4x + 3 = 0

Now you need to do b^2 -4ac, which equals a positive number, so then factorise the equation:

(x-1)(x-3)

so x=1 or x=3

Then substitute these values back into one of the original equations (I suggest y=2x+1 as it is easier) to get the y-coordinates.

(1, 3) and (3,7) as the points where the line intersects the circle.

Hope this helps!

Answered Sat 6th October, 2012 @ 22:16 by Sam

Oops, I totally knew that the equations were that way round. xD

I'm glad to hear the method was at least correct. The mistake was one of those small but generally stupid mistakes that I should have picked up on. :L

Thank you so much for correcting that, it is much appreciated.

Answered Sat 6th October, 2012 @ 22:23 by Bexie