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If two distinct chords of a circle
drawn from a point (p,q) is bisected by x-axis ,then which of the following is correct.a)
friendship is tan 90°.
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hi Nikarita,
EDIT: See next two posts. New diagram in preparation.
Bob
Last edited by Bob (2015-03-04 21:08:11)
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
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Hi Bob;
Should (p,q) be on the circle?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Yes it should. Just realised that myself.
New diagram:
Here the centre is now G. A is still (p,q). Chords are BA and CA. Midpoints E and D respectively. F is the origin.
A and F lie on the circle (and B and C of course!)
Haven't got a method yet, but here's a suggestion:
Introduce letters for the coordinates of B and C.
Work out the coordinates of E and D.
Set their y coordinates to zero and see what happens.
More later.
Bob
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
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Let A be the point (p,q). The circle is
[list=*]
[*]
Clearly the circle passes through the origin O and the point A and has radius OA/2. Thus OA is the diameter the circle.
Because of symmetry we can assume WLOG that p, q > 0 (i.e. A is in the first quadrant). Let a chord drawn from A intersect the x-axis at (t,0). One end of the chord is A; if the axis bisects it, the other end would be (P,Q) where:
[list=1]
[*]
Susbstiting (x,y) = (P,Q) into the equation of the circle and simplifying gives the following quadratic equation in t:
[list=*]
[*]
This equation has two real and distinct solutions in t; therefore its discriminant is strictly positive:
[list=*]
[*]
Last edited by Olinguito (2015-03-04 21:40:50)
Bassaricyon neblina
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Do I get a medal for solving this tough problem?
Bassaricyon neblina
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hi Olinguito
I'm considering the medal but you may have to do a little more to satisfy my limited brain .
As my diagram doesn't have OA as diameter, I'm thinking yours is different to mine. Are you able to post a diagram please?
Bob
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
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Why doesn't your diagram have OA as diameter? The equation of the circle is
[list=*]
[*]
which can be rearranged as
[list=*]
[*]
Does not that make OA the diameter?
Last edited by Olinguito (2015-03-04 22:28:48)
Bassaricyon neblina
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Why? Because I didn't spot that property. Whoops!
I have a diagram. Now I'm happy.
Medal awarded.
Bob
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
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Thanks Bob!
Bassaricyon neblina
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thanks.
and
HAPPY HOLI to all of you.
friendship is tan 90°.
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can you give me a diagram of the double ordinate of a parabola.
I am not able to visualize it.
friendship is tan 90°.
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hi Nikarita
I'm not sure what you mean. Quadratic graphs are parabolas and if you draw a line like y = 5, it will usually cut the graph in two places. Have a look at this:
http://www.mathsisfun.com/data/function … 1=x^2-5x+3
You'll have to enter the line yourself. I cannot find the right syntax to put in the link above directly.
Bob
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
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