Finding where lines and circles meet

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Probability

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Using the equation of the line y = -2x - 2, and the equation x^2 + y^2 + 6x - 8y + 20 = 0

I am looking to find the coordinates of any points at which the circle intersects the line of the first equation above.

This is what I have tried and am stuck at the end hoping that everything else I worked out is OK.


x^2 + y^2 + 6x - 8y + 20 = 0
x^2 + 6x +y^2 - 8y + 20 = 0
(x + 3)^2 + (y - 4)^2 + 20 - 9 - 16 = 0
(x + 3)^2 + (y - 4)^2 = - 5

From;

(x + 3)^2 + (y - 4)^2 = - 5, and using the equation y = -2x - 2, I want to find the coordinates of any points at which the circle intersects the line of this equation here above.

This is what I have tried;

(x + 3)^2 + (y - 4)^2 = - 5


Multiplied out the brackets and simplified to obtain;


x^2 + 6x - 8y^2 + 20 = 0

I think this further simplification in this nest step is correct but unsure;

x + 3x - 4y^2 + 10 = 0

If this is correct, and it will factor I now require to find the factors but am now not sure how to factor this last part!

Any help appreciated.




 
Probability said:
Using the equation of the line y = -2x - 2, and the equation x^2 + y^2 + 6x - 8y + 20 = 0
I am looking to find the coordinates of any points at which the circle intersects the line of the first equation above.
This is what I have tried and am stuck at the end hoping that everything else I worked out is OK.


x^2 + y^2 + 6x - 8y + 20 = 0
x^2 + 6x +y^2 - 8y + 20 = 0
(x + 3)^2 + (y - 4)^2 + 20 - 9 - 16 = 0
(x + 3)^2 + (y - 4)^2 = - 5
Click to expand...
You have a sign error. It should be +5.
 
pka said:
After completing the squares you should have \(\displaystyle (x + 3)^2 + (y - 4)^2=-20+9+16 \)
Click to expand...

Thanks for that, this is a new topic area to me and must say the examples I have seen to date have always been as I wrote the example on here originally.
 
Since the problem is to find where the line y= -2x- 2 and the circle \(\displaystyle x^2+ y^2- 6x- 8y+ 20= 0[/itex] I have no idea why you proceed to complete the square. Replace y in the equation of the circle -2x- 2 and you have \(\displaystyle x^2+ (-2x- 2)^2- 6x- 8(-2x- 2)+ 20= x^2+ 4x^2- 8x+ 4- 6x+ 18x+ 16+ 20= 5x^2+ 4x+40= 0\). Solve that quadratic equation for x, then use y= -2x- 2 to find the corresponding y.\)
 
HallsofIvy said:
Since the problem is to find where the line y= -2x- 2 and the circle \(\displaystyle x^2+ y^2- 6x- 8y+ 20= 0[/itex] I have no idea why you proceed to complete the square. Replace y in the equation of the circle -2x- 2 and you have \(\displaystyle x^2+ (-2x- 2)^2- 6x- 8(-2x- 2)+ 20= x^2+ 4x^2- 8x+ 4- 6x+ 18x+ 16+ 20= 5x^2+ 4x+40= 0\). Solve that quadratic equation for x, then use y= -2x- 2 to find the corresponding y.\)
Click to expand...
\(\displaystyle

Not sure how this method works out?

ax^2 + BX + C

a = 5

B = 4

C = 40

Using the formula I get -784 before I square it?

This indicates there is no roots?

Or maybe I am getting something wrong?\)
 
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