It means if (2,0) is the centre of a circle of radius 5 units,

then there are two points of intersection between the circle and line,

either of which can be labelled P.

Results 1 to 15 of 15

- January 18th 2010, 03:30 AM #1

- Joined
- Dec 2009
- Posts
- 755

- January 18th 2010, 03:57 AM #2

- Joined
- Dec 2009
- Posts
- 3,120
- Thanks
- 1

- January 18th 2010, 04:06 AM #3

- Joined
- Dec 2009
- Posts
- 755

- January 18th 2010, 04:36 AM #4

- Joined
- Dec 2009
- Posts
- 3,120
- Thanks
- 1

How you identify a straight line is by checking the powers of x and y.

If they are both 1, that's a line,

because you can rearrange it to y=mx+c form.

7y=x+23 is

this is a line passing through

with slope

You need to draw a picture (can be a rough sketch) of the axes.

Draw the line roughly.

Take (2,0) as a circle centre and draw a circle of radius 5.

See it now?

- January 18th 2010, 04:44 AM #5

- Joined
- Apr 2005
- Posts
- 16,960
- Thanks
- 2007

I don't understand how you can post the questions you have before and NOT know that any "linear" equation like "7y= x+ 23" has a straight line as graph. As for your question "Do I take the 5units from the x-axis or y-axis?" the answer, as Archie Mead told you, is "neither one"! Since (2, 0) is not on the line 7y= x+ 23, you need to think of the 5 units as defining a circle with center (2, 0) and radius 5. The question is really asking where that circle and line intersect. Do you know what the equation of a circle with center (a, b) and radius r is? If so, write the equation for the circle with center (2, 0) and radius 5 and solve that and 7y= x+ 23 as a pair of simultaneous equations.

If not, you could try to find the point on 7y= x+ 23 closest to (2, 0). A line from (2, 0) to that nearest point will be perpendicular to the line and so will give two right triangles with hypotenuse equal to 5 and one leg equal to the distance from (2,0) to that "nearest" point. You can use the Pythagorean theorem to find the other leg and then use that to find the coordinates of the two points. That will, I suspect, involve a lot harder algebra than the "equation of the circle" method so I suggest you use the equation of the circle.

- January 18th 2010, 04:51 AM #6

- Joined
- Dec 2009
- Posts
- 755

- January 18th 2010, 05:17 AM #7

- Joined
- Apr 2005
- Posts
- 16,960
- Thanks
- 2007

You have already been told, twice, and now a third time, that the two points you want are the two intersections of the line 7y= x+ 23 and the circle with center at (2, 0) and radius 5. I will ask again, do you know the equation of the circle with center at (2, 0) and radius 5?

- January 18th 2010, 05:21 AM #8

- Joined
- Dec 2009
- Posts
- 755

Sorry dude, I have not learnt circles yet... I asked if the line was a curve earlier on because he told me it was a circle and so I was puzzled that a straight line graph could make a circle...

I asked again the second time because I suspected that this was under the topic, circles. I am really sorry about that.

And to your question, I do not know the equation...

- January 18th 2010, 05:58 AM #9

- Joined
- Dec 2009
- Posts
- 3,120
- Thanks
- 1

I didn't tell you that the line was a circle.

I asked if you would draw a sketch with a circle centred at (2,0).

I told you the equation was that of a straight line.

HallsofIvy showed you how to work it out using triangles.

The point (2,0) is on the x-axis.

There are two points on the line for which you wrote the equation, that are 5 units away from (2,0).

If you have a wheel 5 units radius,

isn't every point on the wheel circumference 5 units from the centre ?

The circle is a geometric shape that can be used to visualise the problem.

First make sure you have a picture of the geometry before proceeding.

- January 18th 2010, 06:03 AM #10

- Joined
- Dec 2009
- Posts
- 3,120
- Thanks
- 1

- January 18th 2010, 06:39 AM #11

- Joined
- Dec 2009
- Posts
- 3,120
- Thanks
- 1

- January 18th 2010, 02:59 PM #12

- Joined
- Dec 2009
- Posts
- 3,120
- Thanks
- 1

- January 19th 2010, 04:14 AM #13

- Joined
- Dec 2009
- Posts
- 3,120
- Thanks
- 1

- January 19th 2010, 05:52 AM #14

- Joined
- Dec 2009
- Posts
- 755

Yes, i did some research in my texbook and found some circles formula for centre(x,y). So for the 2 lines to meet, I would have to equate the 2 equation and solve. They work just like any other normal simultaneous equations... They work out fine, thanks for being so concerned

- January 19th 2010, 06:16 AM #15

- Joined
- Dec 2009
- Posts
- 3,120
- Thanks
- 1