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.
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.
this is a line passing through
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?
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.
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...
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.