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The point P is a fixed point on the diameter AB of a circle. Show that for any chord CD of the circle (that is parallel to AB) the equation PC^2 + PD^2 = PA^2 +PB^2. Solve for when point P can be anywhere on the diameter. 

 Feb 5, 2020

Best Answer 

 #1
avatar+18875 
+4

Center this cirle at the origin of a graphing plane.

If the equation of the cirlce is   x2 + y2  =  r2,

the endpoints of the horizonatal diameter through the origin are  A = (-r,0)  and  B = (r,0).

 

Choose the point P to have coordinate of (p,0).

 

Choose the chord parallel to the diameter to have endpoints  C =(-a,b)  and  D = (a,b).

 

Using the distance formula:

     PC  =  sqrt( (p + a)2 + (0 - b)2 )          --->     PC2  =  (p + a)2 + (0 - b)2     --->     p2 + 2ap + a2 + b2 

     PD  =  sqrt( (p - a)2 + (0 - b)2 )          --->     PD2  =  (p - a)2 + (0 - b)2     --->     p2 - 2ap + a2 + b2 

 

----->     PC2 + PD2  =  ( p2 + 2ap + a2 + b2 ) + ( p2 - 2ap + a2 + b2 )  =  2p2 + 2a2 + 2b2

 

Choosing P to be on the radius from the origin to point B:

    PA  =  p + r         and         PB  =  r - p          

 

----->     PA2 + PB2  =  (p + r)2 + (r - p)2  =  ( p2 + 2pr + r) + ( p2 - 2pr + r)  =  2p2 + 2r2

 

But, since the equation for the circle is  x2 + y2  =  r2   and the point  (a,b) is on the circle  a2 + b2  =  r2 

 

2p2 + 2a2 + 2b2     --->     2p2 + 2r2

 

So, they are equal.

 

You can choose P to be on the radius from the origin to point A and get the same result.

 Feb 5, 2020
 #1
avatar+18875 
+4
Best Answer

Center this cirle at the origin of a graphing plane.

If the equation of the cirlce is   x2 + y2  =  r2,

the endpoints of the horizonatal diameter through the origin are  A = (-r,0)  and  B = (r,0).

 

Choose the point P to have coordinate of (p,0).

 

Choose the chord parallel to the diameter to have endpoints  C =(-a,b)  and  D = (a,b).

 

Using the distance formula:

     PC  =  sqrt( (p + a)2 + (0 - b)2 )          --->     PC2  =  (p + a)2 + (0 - b)2     --->     p2 + 2ap + a2 + b2 

     PD  =  sqrt( (p - a)2 + (0 - b)2 )          --->     PD2  =  (p - a)2 + (0 - b)2     --->     p2 - 2ap + a2 + b2 

 

----->     PC2 + PD2  =  ( p2 + 2ap + a2 + b2 ) + ( p2 - 2ap + a2 + b2 )  =  2p2 + 2a2 + 2b2

 

Choosing P to be on the radius from the origin to point B:

    PA  =  p + r         and         PB  =  r - p          

 

----->     PA2 + PB2  =  (p + r)2 + (r - p)2  =  ( p2 + 2pr + r) + ( p2 - 2pr + r)  =  2p2 + 2r2

 

But, since the equation for the circle is  x2 + y2  =  r2   and the point  (a,b) is on the circle  a2 + b2  =  r2 

 

2p2 + 2a2 + 2b2     --->     2p2 + 2r2

 

So, they are equal.

 

You can choose P to be on the radius from the origin to point A and get the same result.

geno3141 Feb 5, 2020
 #2
avatar+109345 
0

Wow, Geno!!!.....that's a very nice proof of this!!!....impressive  !!!!

 

A "Best Answer" fer shure    !!!!!

 

 

cool cool cool

CPhill  Feb 6, 2020
edited by CPhill  Feb 6, 2020

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