Math Is Fun Forum

juantheron

Member

Registered: 2011-10-19

Posts: 312

value of alpha

The values of

for which the point 

  lies in the larger segment of the circle

made by the chord whose eqn is

are

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: value of alpha

hi juantheron

EDIT Don't know if you've read this post yet, but I've made a clearer post now.

Substitute that point (with alphas) into the equation for the circle. 

Simplify to a quadratic and solve for alpha.

Find the points (two answers) and see where they lie in relation to the line.  A sketch will help with this.

Keep any where the point lies where you want.

Arrhh.  Just re-read this.  You want in the circle not on.

Still do what I've said but insert < or > signs in place of the =

I'll refine this later ... got to go out now  ... so I'll check this evening (it's 8.23 am in the UK now)

quick edit:  Yes.  Once you have two values of alpha, replace the = in the circle equation with a < (to be inside the circle)

That leads to a range of values. 

Also substitute the alpha terms into the equation for the line, which will lead to a second inequality.  Put the two inequalities together for the final answer.

Bob

Last edited by Bob (2011-10-24 00:16:55)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: value of alpha

OK. improved answer coming up.

I'm going to use 'a' rather than 'alpha' to save time typing.

(i)  Substitute x = a - 1 and  y = a + 1   in the circle equation.

(ii) Simplify to a quadratic and get two values of a.

These are the two points on the circle.

(iii)  But the question asks for in the circle.

For the general equation of a circle

centre (p,q) and radius r

If a pair of coordinates (x,y) fit this equation the point lies on the edge of the circle because the distance from the centre to the point equals the radius.

If

it means the distance from centre to the point is less than the radius so the point is inside the circle.

You should have got to a quadratic (a-b)(a+c) = 0 in part (ii)

So what we need to do is look at the inequality

(a-b)(a+c) < 0

If two numbers multiply to give a negative result one must be positive and one must be negative. 

So you'll need to consider

a - b > 0 AND a + c < 0.  You should be able to see this is impossible!

OR

a - b < 0 AND a + c > 0.  This case will be possible giving a range of values for a as follows

-c < a < b

That deals with inside the circle.

(iv)  Now look at the sketch graph below.  I have NOT computed the radius accurately, but the sketch should be sufficient to show that the major segment is left and below the line.

So put x = a - 1 and y = a + 1 in the equation of the line.

The values of a will tell you about a point on the line.  To be below and left of the line, change the = 2 into < 2

This will give a second inequality for points left and below the line.

(v) Finally, put the two inequalities together to make a single inequality for all points satisfying both constraints.  That's your final answer.

Hope that helps, 

Bob

Last edited by Bob (2011-10-24 00:42:30)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob