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How to prove point lies in/outside circle? (C2)

Hi -

Please can someone tell me how I can determine whether a point lies inside or outside a circle (with the equation of the circle?)

Thanks

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Put the x coordinate of the point into the equation and show that it is not equal to the y value of the point you have been given?
John Locke
Put the x coordinate of the point into the equation and show that it is not equal to the y value of the point you have been given?

I don't understand lol...
Reply 3
Avatar for 0-)
0-)
11
Eq of circle ---- x2+y2=r2x^2+y^2=r^2.

If for a point P=(a,b)P=(a,b),

a2+b2≤r2a^2+b^2 \leq r^2

then P lies in (or on the boundary of) the circle.
Reply 4
Avatar for Hrov
Hrov
2
i think its something to do with the radius.
so find the centre, and if the point to the centre is greater in length (of the radius) then the point lays outside of the circle
Darkest Knight
I don't understand lol...


Woops, my bad. Misread the question.
0-)
Eq of circle ---- x2+y2=r2x^2+y^2=r^2.

If for a point P=(x1,y1)P=(x_1,y_1),

x12+y12≤r2x_1^2+y_1^2 \leq r^2

then P lies in (or on the boundary of) the circle.

Thanks, i don't undersatnd the x (2/1) + y (2/1) if you could explain that for me please?

Bit confused too, as seem to be different methods?
Reply 7
Avatar for jamesh_91
jamesh_91
1
that 2 1 is just the subscript of using the point x1 so its the same as the first equation of the circle just using the co-ordinates out of the point p
Reply 8
Avatar for 0-)
0-)
11
Darkest Knight
Thanks, i don't undersatnd the x (2/1) + y (2/1) if you could explain that for me please?

Bit confused too, as seem to be different methods?

I edited my post to make it less confusing. Can you understand it now?
Reply 9
Avatar for 0-)
0-)
11
Oops. I shouldn't have assumed the circle was centred at the origin.
thanks very much, your method is differnet to 0- )'s right (can I use his/hers too)?

also, if i'm given the equation of a circle, and it has = 4 at the end for example.. does that mean the radius is 4, or it is 4^2 ?

Thanks
thanks, equaton is

(x+3)^2 + (y-1)^2 = 4
Thanks, so in that particular question... with the point being (-2,3). As 13>2 , it is outside of the circle?

I'd appreciate if you could also help me on another question too lol.

Equation of a circle is: x^2 + y^2 + 6x -4y +4 = 0, I need to prove that the line y = x + 5 passes through it.. and have no idea how!

Thanks
not sure if I did that right.. but I end p with x^2 + 6x +0.5 = 0 .. is that right?!

Thanks
okay, the equation of circle, x^2 + y^2 + 6x -4y +4 = 0, after completing the square is (x+3)^2 + (y-2)^2 = 17
Expand to get 2x^2 + 12x + 18 = 17
divide by 2: x^2 + 6x + 9 = 8.5
simplify x^2 + 6x + 0.5 = 0 ?
is that okay?

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