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Let l be a line and C be a circle.

$y=x+d$, where $d$ is to be determined.

$C=x^2+y^2=4$

  • Pick a value for $d$ so that l and C intersect at one point.
  • Pick a value for $d$ so that l and C intersect at two distinct points.
  • Pick a value for $d$ so that l and C do not intersect.

Am I able to use this: How do I calculate the intersection(s) of a straight line and a circle??

Graphically, I could answer B and C. I'm hoping there's an algebraic way to solve all three parts.

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Information


We have two equations:

  • $ y = x + d $

  • $ x^2 + y^2 = 4 $

We must find the number of intersections depending on $d$.


Solution


Step 1

Let us begin with finding the relation between $x$ and $d$.

Why? We need to see how we need to manipulate $d$ to get 0, 1, or 2 valid $x$ values. Each valid $x$ value will (obviously) correspond to a point of intersection.

$ x^2 + y^2 = 4 $

$ x^2 + (x + d)^2 = 4 $

$ (2)x^2 + (2d)x + (d^2 - 4) = 0 $

$ x = \frac{-(2d) \pm \sqrt{(2d)^2 - 4(2)(d^2 - 4)}}{2(2)} = \frac{-d \pm \sqrt{8 - d^2}}{2} $


Step 2

Now we shall analyze our relation between $x$ and $d$.

Let $ x = \frac{-d \pm \sqrt{A}}{2} $ where $ A = 8 - d^2 $.


Step 2A

If $ A = 0 $ then $x$ has one solution.

What must $d$ be such that $A = 0$?

$ A = 8 - d^2 = 0 $

$ d^2 = 8 $

$ d = \pm 2\sqrt{2} $


Step 2B

If $ A > 0 $ then $x$ has 2 solutions.

What must $d$ be such that $A > 0$?

$ A = 8 - d^2 > 0 $

$ d^2 < 8 $

$ d < \pm 2\sqrt{2} $

$ -2\sqrt{2} < d < 2\sqrt{2} $


Step 2C

If $ A < 0 $ then $x$ has 0 solutions.

What must $d$ be such that $A < 0$?

$ A = 8 - d^2 < 0 $

$ d^2 > 8 $

$ d > \pm 2\sqrt{2} $

$ d < -2\sqrt{2} $ or $ 2\sqrt{2} < d $


Answer

  • 0 solutions: $ d < -2\sqrt{2} $ or $ 2\sqrt{2} < d $
  • 1 solution: $ d = \pm 2\sqrt{2} $
  • 2 solutions: $ -2\sqrt{2} < d < 2\sqrt{2} $
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  • $\begingroup$ Thank you, Sir Jony, this was a great explanation! I appreciate the step-by-step work you showed, I understand perfectly now! $\endgroup$ – jbrow35 Jun 24 '16 at 22:09
  • $\begingroup$ @jbrow35 -- No, thank you for asking such an interesting question. A detailed (yet easy-to-read) explanation is my signature answer. :) $\endgroup$ – Fine Man Jun 24 '16 at 22:41
  • $\begingroup$ Typos in your answer last 3 lines? $\endgroup$ – Narasimham Jun 25 '16 at 15:35
  • $\begingroup$ @Narasimham -- Oops, I misordered them. Thanks! $\endgroup$ – Fine Man Jun 25 '16 at 18:00
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Substitute $y=x+d$ into the equation and you will get $x^2 + (x+d)^2 = 4$. This is a quadratic equation. When $D>0$ the line intersects the cirlce. When $D=0$ then they are tangent to each other and when $D<0$ then they don't intersect at all.

Use this to determine for which values of $d$ the determinant $D$ obtaints the above-mentioned values.

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  • $\begingroup$ Thanks for replying so quickly, Stefan4024! I appreciate the help! $\endgroup$ – jbrow35 Jun 24 '16 at 22:15
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Well, I don't know about algorithm, however for such problems I suggest you to just plot and understand what's going on. For example you know that the circle $C$ has radius 2 and is centred at the origin. While the line $y = x + d$ is parallel to the bisecant of the first and third quadrant in the carthesian plane.

Hence by simply taking $y = x$, you intersect the circle in two points. In general, to find all the values $d$ that would give you this result you need to find where the line intersects the circle in one point. This happens when the line is TANGENT to the circle. How do you find this? well it's pretty simple, just put $y=x + d$ into the equation of the circle, to get: $x^2 + (x+d)^2 = 4$. Now use algebra and get $2x^2 + 2dx+d^2-4 = 0$.

From this you can use the standard formula and obtain: \begin{equation} \frac{-2d \pm\sqrt{4d^2-8(d^2-4)}}{4} \end{equation} Now, the standard formula for $ax^2+bx+c=0$ would be $\frac{-b \pm\sqrt{b^2-4ac}}{2a}$. You can see that part of it is underneath the square root, hence it MUST be POSITIVE.

This thing has a name, it's called the determinant, often denoted by $\Delta = b^2 -4ac$. You can see that if this is negative, the quadratic formula has no solutions. If it is equals to zero, then you have one solution (well, actually you have two, but they are the same number). If it is positive, then you have two solutions.

When does this happen? $\Delta = 0$ when $b^2 = 4ac$, $\Delta > 0$ when $b^2 > 4ac$ and $\Delta <0$ when $b^2 < 4ac$.

In your case $\Delta = 4d^2-8(d^2-4) = -4d^2+32$, hence you have:

  1. $\Delta = 0 \rightarrow d^2 = 8 \rightarrow\,\, d = \pm 2\sqrt{2}$
  2. $\Delta > 0 \rightarrow d^2 < 8 \rightarrow \,\,2\sqrt{2}< d < 2\sqrt{2}$
  3. $\Delta < 0 \rightarrow d^2 > 8 \rightarrow\,\, d > 2\sqrt{2\,}$ or $\,d < -2\sqrt{2}$

What does this tell us? Well when the discriminant is positive, you have two solutions, so you have two intersections. When it is equal to zero, you have one point of intersection and when it is negative, you have no point of intersection.

So if you take either $d = \pm 2\sqrt{2}$, you'll get one intersection, if you get $-2\sqrt{2} < d < 2\sqrt{2}$, then you have two points of intersection, while if $d > 2\sqrt{2}$ or $d < -2\sqrt{2}$, then you have no points of intersection.

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  • $\begingroup$ Thank you for the help, Euler_Salter, I appreciate the thorough explanation! $\endgroup$ – jbrow35 Jun 24 '16 at 22:14
  • $\begingroup$ @jbrow35, don't worry $\endgroup$ – Euler_Salter Jun 24 '16 at 23:00
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To find the intersection of circle $x^2+ y^2= 4$ and the line $y= x+ d$ means to find a point (x, y) that satisfy both equations. In particular, the "y" in both equations must satisfy y= x+ d. Putting that into the circle equation $x^2+ y^2= x^2+ (x+ d)^2= 2x^2+ 2dx+ d^2= 4$.

That is a quadratic equation so there may be two, one or no solution: for any quadratic equation $ax^2+ bx+ c= 0$, solutions are of the form $x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$ (that's the "quadratic formula). If the "discriminant", the "$b^2- 4ac$" inside the square root, is positive there are two solutons using + and -. If it is 0, there is only one solution, if it is negative, there is no solution.

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  • $\begingroup$ Thank you for your help, user247327! $\endgroup$ – jbrow35 Jun 24 '16 at 22:14
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For intersection points both equations $$ y = x + d \\ x^2 + y^2 = 2^2 $$ must hold. Inserting the first into the second we get $$ y = x + d $$ and $$ 2^2 = x^2 + (x+ d)^2 = 2 x^2 + 2xd + d^2 \iff \\ 2 = x^2 + x d + d^2/2 = (x + d/2)^2 - d^2/4 + d^2/2 \iff \\ 2 - d^2/4 = (x + d/2)^2 \iff \\ x = \frac{-d \pm \sqrt{8 - d^2}}{2} $$ Which gives the solutions $$ P = \left( \frac{-d \pm \sqrt{8 - d^2}}{2}, \frac{-d \pm \sqrt{8 - d^2}}{2} + d \right) $$ The number of real solutions is determined by the argument of the square root $$ \Delta(d) = 8 - d^2 $$ If $\Delta(d) = 0$ or $d = \pm\sqrt{8} = \pm 2\sqrt{2}$ there is one solution.
If $\Delta(d) > 0$ or $\lvert d \rvert < 2\sqrt{2}$ there are two solutions.
If $\Delta(d) < 0$ or $\lvert d \rvert > 2\sqrt{2}$ there is no (real) solution.

image

You can fiddle with it here.

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  • $\begingroup$ Thank you, mvw, I appreciate the explanation...especially the interactive graph you linked to! :) $\endgroup$ – jbrow35 Jun 24 '16 at 22:11
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The equations of straight line and circle.

$$ y = x + d ; \; x^2 + y^2 = 4 $$

Eliminate $y$ , simplify

$$ x^2 + x d + (d^2/2-2) $$

Changing $d$ means raising or lowering the line to get 2 different, 2 coincident, null points of intersection.

Zero determinant is condition for tangentcy, positive for two real points, negative for two complex roots, without any real cutting.

For first condition

$$ d^2 = 4\cdot 1\cdot(d^2/2-2) $$

or $$ d_{tangent}= 2 \sqrt2 $$

for tangentcy. Thus we have

$$ \pm d = d_{tangent}, $$

$$ d< -d_{tangent} , d > d_{tangent} $$

$$ -d_{tangent} < d < d_{tangent} $$

for the 5 cases..

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  • $\begingroup$ Great, thank you for your help, Narasimham! $\endgroup$ – jbrow35 Jun 24 '16 at 22:12

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