# Questions about Lines and Circles. Lots of them!

I know I am supposed to ask a specific question, but there's just too many that I would have to ask [it would be like spam] since I missed one week of school because of a family thing and we have an exam this Tuesday, and the teacher's got Mondays off, since it's only 'practice' classes, which means I can't ask her. So, I will just group them here, hoping someone will answer.

# Lines

### Explicit, Implicit and Segment Line equation

Let's say you have this line equation (implicit form): $x-2y-3=0$

How to convert that (back and forth) into explicit and segment forms.

### Common point / Line crossing point

You have two lines: $x-2y-3=0$ and $3x-2y-1=0$

How do you determine where they cross (and if they cross) [This might be a bad example].

### Angles between lines

So, taking the two lines from the above example: $x-2y-3=0$ and $3x-2y-1=0$

How would you determine the angle between them (if they're not parallel, that is).

### $k$ - the direction coefficient

When given the following line equation: $3x-2y-1=0$. How does one calculate $k$?

# Circles

### Writing the 'proper' circle equation

I know, the title is a bit... odd, but I will provide an example.

Let's say you're given this circle equation: $x^2+y^2+6x-2y=0$

That has to be transformed into something that resembles: $(x-p)^2+(y-q)^2=r^2$

I would take this approach: $x^2+y^2+6x-2y=0$ / $+3^2-3^2+1-1$

When sorted out you get: $x^2+6x+3^2+y^2-2y+1=8$ which is in fact: $(x+3)^2+(y-1)^2=8$. I hope I'm right! :P

### Defining whether a point is a part of the circle

Let's have we have this circle: $(x+3)^2+(y-1)^2=8$, how would you define whether point $T$ is a part of the circle's 'ring.' I'm going on a limb here, and I'll just point out a thought: Would you just replace the $x$ and $y$ in the equation with the coordinates of $T$?

### Tangent

This one's a little tougher (at least I think so). So, you have $(x+3)^2+(y-1)^2=8$ and a point $T(-2,4)$ which can be on or off the circle. Now, I know there's 2 approaches: One if the point is on the circle and the second one if it's off it. So, you have to write a Line equation of the Tangent that goes through that point. I really couldn't figure this one at all I have a vague idea of how to do all the above mentioned, but this one's a bit a total mess.

### Circle equation of a circle that touches both of the axis and the circle's centre point lies on a given line

Whew, that took a while to compose... Well, Let's say we have the line $x-2y-6=0$ and we have to determine the centre and the equation of the circle, taking into consideration that the circle touches both axis. The only thing I can gather from that is that $|q|=|p|=r$

Well, I hope someone actually reads and answers this, because I've been writing it for the past hour flipping through the textbook like a madman. And it would save my life.

## Thanks!

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Ok: so these feel lot classwork, and it would be best if you showed some of your own work. This is for two reasons: it shows us that you aren't using us just for answers, and it shows us the methods that you are allowed to use. A quick question: do you know dot products or anything of vectors? And if these are homework, please say so. – mixedmath Jun 5 '11 at 14:06
Now, 2 clarifications: [1] What do you mean by explicit, implicit, and segmented forms? [2] Is a k-direction coefficient the same thing as 'slope?' – mixedmath Jun 5 '11 at 14:15
@mixedmath Well, we did vectors a while back, but I don't think we're allowed to use them. If it were homework I would've told you so :.). I can show you some examples of what they did while I was gone. Well, actually the "writing proper circle equation" was a part of this weeks homework. I'll edit question and let you know in the comments when I'm done. – omninonsense Jun 5 '11 at 14:16
Observe that $x-2y-3=0\Leftrightarrow y=\frac{1}{2}x-\frac{3}{2}$. – Américo Tavares Jun 5 '11 at 14:53

This is my contribution. Observe that all these equations are equivalent

$$x-2y-3=0\Leftrightarrow y=\frac{1}{2}x-\frac{3}{2}\Leftrightarrow \frac{x}{3}+\frac{y}{-3/2}=1.$$

And the same idea applies to the general equation $Ax+Bx+C=0$.

Lines Common point. Suppose you have two straight lines with equations

$$Ax+By+C=0$$

$$A^{\prime }x+B^{\prime }y+C^{\prime }=0.$$

They are not parallel if and only if $AB^{\prime }-A^{\prime }B\neq 0$ (or equivalently $\frac{B^{\prime }}{B}\neq \frac{A^{\prime }}{A}$)

Since for the lines with equations

$$x-2y-3=0$$

$$3x-2y-1=0$$

we have $\frac{2}{2}\neq \frac{3}{1}$, they cross each other. To find the coordinates $(x,y)$ of the intersecting point you have to solve the system of equations. For instance as follows

$$\left\{ \begin{array}{c} x-2y-3=0 \\ 3x-2y-1=0% \end{array}% \right. \Leftrightarrow \left\{ \begin{array}{c} x-2y-3=0 \\ -3x+2y+1=0% \end{array}% \right.$$

$$\Leftrightarrow \left\{ \begin{array}{c} x-2y-3=0 \\ -2x-2=0% \end{array}% \right. \Leftrightarrow \left\{ \begin{array}{c} -1-2y-3=0 \\ x=-1% \end{array}% \right. \Leftrightarrow \left\{ \begin{array}{c} y=-2 \\ x=-1% \end{array}% \right.$$

The second system results from the first by multiplying the second equation by $-1$. The third, by replacing the second equation by the sum of both equations.

If you solve with matrices

$$% \begin{bmatrix} A & B \\ A^{\prime } & B^{\prime }% \end{bmatrix}% \begin{bmatrix} x \\ y% \end{bmatrix}% =% \begin{bmatrix} -C \\ -C^{\prime }% \end{bmatrix}%$$

the expression $AB^{\prime }-A^{\prime }B$ is the determinant of the matrix formed by the coefficients of $x$ and $y$

$$\begin{bmatrix} A & B \\ A^{\prime } & B^{\prime }% \end{bmatrix}$$

Angles between lines. The smallest angle $0\leq \theta \leq \pi /2$ between two lines is such that

$$\tan \theta =\left\vert \frac{m-m^{\prime }}{1+mm^{\prime }}\right\vert ,$$

where $m$ is the slope of the line with equation $Ax+By+C=0$ and $m^{\prime }$ is the sope of $A^{\prime }x+B^{\prime }y+C^{\prime }=0$.

This result is derived from the trigonometric identity

$$\tan (\alpha -\alpha')=\frac{\tan \alpha-\tan \alpha'}{1+\tan \alpha\cdot\tan \alpha'}$$

where $\alpha,\alpha'$ are the angles of inclination of the first and the second line, respectivelly. And $m=\tan \alpha,m'=\tan \alpha'$.

The other angle between these two lines is $\pi -\theta$.

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Okay, I understand most of what you've said, I'm just confused about one thing. Is the following correct?: $Ax-By-C=0\Leftrightarrow y=\frac{A}{B}x-\frac{C}{B}\Leftrightarrow \frac{x}{C}+\frac{y}{-C/B}=1.$ I am assuming I got it wrong... – omninonsense Jun 5 '11 at 16:54
Uh, it looks a little broken, but I think it's readable :x Sorry about that. – omninonsense Jun 5 '11 at 16:55
Yes. $$Ax-By-C=0\Leftrightarrow \frac{x}{C/A}+\frac{y}{-C/B}=1$$ or in the usual form $$Ax+By+C=0\Leftrightarrow \frac{x}{-C/A}+\frac{y}{-C/B}=1$$ – Américo Tavares Jun 5 '11 at 17:25

I read through some of your circle questions. The first three go together I believe, so here are some tips. For the first one, you are correct in adding some constants to get the correct form. You have \begin{align*} x^2+y^2+6x-2y=0 &\implies x^2+6x+y^2-2y=0\\ &\implies x^2+6x+3^2+y^2-2y+1^2=3^2+1^2\\ &\implies (x+3)^2+(y-1)^2=10 \end{align*} so I believe $r=\sqrt{10}$, not $\sqrt{8}$. This makes the third problem easier, as you will see.

For the second problem, just plug in a given point. If it satisfies the equation, that means the point does lie on the circumference of the circle, since this is precisely what the equation describes.

Now for your third problem, I assume the circle is actually given by $(x+3)^2+(y-1)^2=10$. So by the second problem $T(-2,4)$ is on the circle. Now the tangent through that point is perpendicular to the line through the center of the circle, $(-3,1)$ and the given point $(-2,4)$. Do you recall how to find the slope of a line? Once you have that, remember that the slope of two perpendicular lines have a product of $-1$. You can then use point-slope form to find the equation of the desired tangent line.

Added: I took a look at the fourth problem as well. Let $(x,y)$ be the center of your desired circle. It is on the given line, so solving for $y$, your center is $(x,(x-6)/2)$. Now if the circle is to be tangent to both the axes, you want the projections from the center of the circle to be of equal distance. The projections onto the $x$-axis and $y$-axis are just $(x,0)$ and $(0,y)=(0,(x-6)/2)$, respectively. (If it's not familiar to you, a projection here just means that you drop a point perpendicularly onto one of the axes.)

So the distances of these two projections are just $|x|$ and $|(x-6)/2|$. If you solve this equation, you find the $x$-coordinate of your center, from which you can find the $y$-coordinate. You also have radius $|x|$. By the way, I believe there are two such possible circles.

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Oh God. Wow, you're a genius! Side note: I don't know how to calculate a slope. D: [Frankly, I've been calling it ~direction~ this whole time. Well, I'm not English, so I suppose it's an understandable mistake.] Thanks so much – omninonsense Jun 5 '11 at 14:24
Sure, you can calculate the slope $m$ as $m=(y_2-y_1)/(x_2-x_1)$, where $(x_1,x_2)$ and $(y_1,y_2)$ are two points on the line. I hope it helps. – yunone Jun 5 '11 at 14:33

To determine the intersection (crossing) of two different lines, you set them equal to each other and you solve. So you would solve the system of equations (in your example)

\begin{align} x-2y-3 &= 0 \\ 3x - 2y - 1 &= 0 \end{align}

There are many ways to determine the angle between two lines. You might find their slopes and then use trig (arctangent, for example). You might use dot products, if that's what you were taught. But you usually just make a triangle from the two slopes and solve for the angle (as you'll have all the side lengths).

Finally, if $\kappa$ is what you call 'slope,' then you look for the coefficient m when the equation is in the form $y = mx + b$.

Other than that, yunone's answer solves the circles.

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