# Thread: Coordinates of intersection of a line and circle?

1. ## Coordinates of intersection of a line and circle?

How do you get it? I've tried for 30 minutes now...

y=x-3
x^2+y^2=65

Yes I do know substitution, plus add, and the third way to find intersections.

2. Why don't you show us what you've done and we'll see where you went wrong?

3. Originally Posted by o_O
Why don't you show us what you've done and we'll see where you went wrong?
This is one of the attempts I made that I could find on paper.

I substituted (x-3)^2 for y^2
x^2 + (x-3)^2 = 65
x^2 + x^2 - 6s + 9 -9 = 65 -9
(2x^2-6x)2 = 56/2
x^2-3x=28
x(2x-3)=28

4. Not a bad effort. Now, you don't want to end up with your last line unless you have a 0 on the right hand side as that allows you to equate each factor equal to 0 which we don't have here.

What you should do is try to get your expression in the form of a quadratic ( $ax^{2} + bx + c = {\color{red}0}$). The 0 is important as this is the key for using the quadratic formula or to factor for the reason mentioned above:
$x^{2} + x^2 - 6x + 9 = 65$
$2x^{2} - 6x - 56=0$
$x^{2} - 3x - 28 = 0$ (divided both sides by 2)

Can you see what to do now?

5. Originally Posted by o_O
Not a bad effort. Now, you don't want to end up with your last line unless you have a 0 on the right hand side as that allows you to equate each factor equal to 0 which we don't have here.

What you should do is try to get your expression in the form of a quadratic ( $ax^{2} + bx + c = {\color{red}0}$). The 0 is important as this is the key for using the quadratic formula or to factor for the reason mentioned above:
$x^{2} + x^2 - 6x + 9 = 65$
$2x^{2} - 6x - 56=0$
$x^{2} - 3x - 28 = 0$ (divided both sides by 2)

Can you see what to do now?
Thanks!

Did you get (7,4) and (-4,7), too?

6. Originally Posted by AlphaRock
I substituted (x-3)^2 for y^2
x^2 + (x-3)^2 = 65
x^2 + x^2 - 6s + 9 -9 = 65 -9
(2x^2-6x)2 = 56/2
x^2-3x=28
x(2x-3)=28
Hello, AlphaRock! Your work looks fine so far (except for the 2 in the last step). Now let us solve it!

Are you familiar with solving quadratic equations? There are a few methods you can use:

First, you can try factoring:

$x^2-3x=28\Rightarrow x^2-3x-28=0$

$\Rightarrow x^2+(4-7)x+(-7)\cdot4=0$

$\Rightarrow (x+4)(x-7)=0$

Next, if you can't figure out how to factor the expression, or if the expression is simply too messy to easily factor, you can use the quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a, b,\text{ and }c$ are the coefficients of the 2nd-degree, 1st-degree, and constant terms, respectively.

In this case, we have
$x^2-3x=28\Rightarrow x^2-3x-28=ax^2+bx+c=0$

So, in the formula, $a=1, b=-3,\text{ and }c=-28$.

Alternatively, you can complete the square. For example:

$x^2-3x=28\Rightarrow x^2-3x+\frac94=28+\frac94$

$\Rightarrow\left(x-\frac32\right)^2=\frac{121}4$

$\Rightarrow x-\frac32=\pm\sqrt{\frac{121}4}$

$\Rightarrow x=\frac32\pm\frac{11}2$

I hope that helps!

Edit: Beaten by o_O! I'm too slow with my LaTeX!

7. Looks good.

8. Ah, here's a more harder question:

4x+y+11=0 and (x-5)^2+(y-3)^2=16

What do I do to solve this question...? I'm clueless on what ot do next.

Here's what I did:

y=(-11-4x)

(x-5)^2 + (-11-4x-3)^2 = 16
x^2-10x+25+(-14-4x)^2 = 16
x^2-10x+25 + 196 + 16x^2 + 112x = 16
17x^2+102x+221-16 = 16-16
17x^2+102x+205=0

>>Imaginary. What do I do...? THis is all my work.

9. Originally Posted by AlphaRock
4x+y+11=0 and (x-5)^2+(y-3)^2=16

What do I do to solve this question...? I'm clueless on what ot do next.

Here's what I did:

y=(-11-4x)

(x-5)^2 + (-11-4x-3)^2 = 16
x^2-10x+25+(-14-4x)^2 = 16
x^2-10x+25 + 196 + 16x^2 + 112x = 16
17x^2+102x+221-16 = 16-16
17x^2+102x+205=0

>>Imaginary. What do I do...? THis is all my work.
Your work is okay so far. If you continue, using the quadratic formula, you should get

$x=-\frac{2\sqrt{221}\,\text{i}+51}{17},\ \frac{2\sqrt{221}\,\text{i}-51}{17};\
y=\frac{8\sqrt{221}\,\text{i}+17}{17},\ -\frac{8\sqrt{221}\,\text{i}-17}{17}$

Edit: But, make sure you have the right problem, because I notice that if you change the addition in the second equation to a subtraction the two graphs will intersect.