IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

I am sure to be drowned in answers, yet I am curious:

A circle is drawn around the point (0,

2). Prove that there are at most 2 rational points on the circle [rational points means both x and y are rational]

amangupta_1991

i m not sure about the answer, but here it goes
a genral eqn of a circle around 0,root 2
is x^2+y^2-(root2/2)y +c=0
this eqn is a quadratic in y^2 in which a=1, b=-root2/2 c=x^2+c , i have no idea how to solve after this point ..lol

hsbhatt

The reason I posed this qn is that I saw a ridiculous answer somewhere in this formum I can't recall where. I thought that since it is a previous JEE prob, it would have been discussed already and was curious to see the solns

I'll wait one day b4 posting

konichiwa2x

Consider an equilaterial triangle with its vertices on the circle. The centroid of the triangle is

.

and

. RHS is irrational, hence atleast one of the variables on the LHS have to be irrational. => at most two are rational.

rohith291991

y equilateral? any circle wud do....

hsbhatt

Konichiwa,

How do you prove that in the entire circle, there are at two rational points and no more. Any number of equilateral triangles could be drawn and we cud conclude that each has at the most two rational points.

Also see if it is possible to tell under what condition there are no rational points at all

Now on to you

hsbhatt

The equation of the family of circles with radius (0,

2) is

x²+(y-

2)² = r².

It is easily proved that if r is rational then there are no rational points.

Now, with r irrational, let's assume that we have at least one rational point (x₁,y₁) on the circle. Now, suppose (x₂,y₂) is another such point, we must have

x₁²+(y₁-

2)² = x₂²+(y₂-

2)².

Simplifying we get

(x₁²+y₁²-x₂²-y₂²) + 2

2(y₂-y₁) = 0.

we know that if a+b

2 = 0 with a,b rational then a=0 and b= 0.

This gives y₁ = y₂. Correspondingly x₂ = -x₁. Hence, there is only one other rational point which is (-x₁,y₁). Hence, if rational points exist there are atmost 2 such points.

konichiwa2x

thanks! I should have spent more time on it.

And for your 2nd question, i.e the condition that a circle will not contain any rational point on its circumference is that both x-coordinate and y-coordinate of the centre of the circle must be irrational and such a circle should not pass through the origin.

Let us assume all the points on the circle are irrational.
The family of all circles with centre

is

Let

and

be two such rational points on the circumferrence of the circle.
Hence,

Quite clearly, this is not possible for possible for any rational

except

.

hsbhatt

actually I meant to ask in the context of the problem, are there circles with center (0,

2) which is answered in my solution - when the radius is a rational number.

konichiwa2x

oh ok...i understand...didnt see that part about 'r' being rational at first.

konichiwa2x

and here is another approach...

If the circle has radius r then its equation is

, or

. If x and y are both rational then r^2 must be of the form

,(failed to see this point first time) where a and b are both rational. Then

. This can only have a rational solution if

. The straight line

meets the circle in at most two points.

hsbhatt

yes. this is quite neat.

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