Question 554469
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(a) the sum of the distance from a point P to (4,0) and (-4,0) is 9. if the abscissa of P is 1, find its ordinate.
(b) the center of a circle is at (-3,-2). if a chord of length 4 is bisected at (3,1), find the length of the radius.
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        I am @ikleyn.   Hello again.

        This  'n2'  is my second nickname,  which I created to place here my solution to  part  (a)  of the problem.


        Tutor @greenestamps solved this part nicely by applying theory of ellipses.

        In my solution below,  I work in the frame of knowledge related to triangle geometry,  only.



<pre>
Let A, C and P be the given points A = (-4,0), C = (4,0).
Let P = (1,y) be the point, for which we want to find its coordinate 'y' such that

    |AP| + |CP| = 9.    (1)


Notice that line AC is horizontal (lies on x-axis of the (x,y) coordinate system).


Draw a perpendicular PD from vertex P of triangle APC to its base AC, 
so point D = (1,0) is the intersection of the perpendicular with AC.


Thus we have now triangle APC and two right-angled triangles ADP and CDP.

Let 'a' be the length CP and 'c' be the length AP:  

    a = |CP|,  c = |AP|.    (2)


We have  |AD| = 1 - (-4) = 5;  |CD| = 4 - 1 = 3.   (3)


Perpendicular PD is the common leg of triangles ADP and CDP, so we can write, using Pythagorean equation 

   |AP|^2 - |AD|^2 = |CP|^2 - |CD|^2.



Substituting here from relations (2) and (3), we get this equation

   c^2 - 5^2 = a^2 - 3^2,


which implies 

    c^2 - a^2 = 5^2 - 3^2,

    c^2 - a^2 = 16.    (4)



So, now we have this system of equations (1) and (4)

    c   + a   =  9.    (1')

    c^2 - a^2 = 16,    (4')


Now we are on a finish line to complete the solution.



In equation (4'),  factor left side as  c^2 - a^2 = (c+a)*(c-a) and replace (c+a) by 9,
based on equation (1').  Then instead the system (1'), (4') you will get the system

    a  + c =  9.    (1'')

    9(c-a) = 16.    (4'')



Open parentheses in (4'')  and multiply equation (1'')  by 9  (both sides)

    9c + 9a = 81.

    9c - 9a = 16,


Add      the two last equations and get  18c = 81 + 16 = 97,  c = 97/18.

Subtract the two last equations and get  18a = 81 - 16 = 65,  c = 65/18.



Now we can find 'y'

    y^2 = |PD|^2 = |CP|^2 - |AD|^2 = a^2 - 3^2 = {{{(65/18)^2 - 9}}} = {{{(65^2-18^2*9)/18^2}}} = {{{1309/18^2}}},

    y = {{{sqrt(1309)/18}}} = 2.010005835,  or  y = 2.01, approximately.


<U>ANSWER</U>.  y = {{{sqrt(1309)/18}}} = 2.010005835,  or  y = 2.01, approximately.
</pre>

Solved.