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The points (0,4) and (1,8) lie on a circle whose center is on the x-axis. What is the radius of the circle?

 May 15, 2021
 #1
avatar+311 
0

r = sqrt(6497) / 16

 

wink

 May 15, 2021
 #2
avatar+22082 
-1

The center of the circle is on the perpendicular bisector of each chord of the circle.

 

The plan:

1)  find the equation of the perpendicular bisector of the chord whose endpoints are  (0, 4)  and  (1, 8),

2)  find where this equation crosses the x-axis; this will be the center.

3)  find the distance from this point (the center) to the point  (0, 4).

 

1)  The midpoint of  (0, 4)  and  (1, 8)  is  ( ½, 6).

      The slope of the chord is:  m  =  (8 - 4) / (1 - 0)  =  4.

      The perpendicular will have a slope of  - ¼.

      The equation of this line is:  y - 6  =  - ¼( x - ½ ).

      Multiplying by -4:     -4y + 24  =  x - ½

      Multiplying by 2:       -8y  + 48  =  2x - 1

 

2)  On the x-axis, the value of y is zero.

     Making this subsitution:  -8(0) + 48  =  2x - 1

                                                        49  =  2x     --->     x  =  24.5

      So, the center is  (24.5, 0).

 

3)  Finding the distance from  (24.5, 0) to (0, 4):

     distance  =  sqrt(  (4 - 0)2  +  (0 - 24.5)2  )     =     sqrt( 16 + 600.25  )     =     sqrt(  616.25  )

 May 15, 2021
 #3
avatar+32762 
-1

Here is a second way:

the two given points are equidistant from the center    (x, 0)

Use distance formulas for the two points and the center

(x-0)^2  + 4^2 = (x-1)^2 + 8^2

   solve for x = 49/2      so circle center is   49/2 , 0

 

use one pont to solve for r in the circle formula   (0,4 )

( 0 - 49/2)^2 + (4-0)^2 = r^2

   r^2 = 616.25                       r = sqrt (616.25)

 May 15, 2021
 #4
avatar+32762 
-1

Here is a desmos picture:

 

 May 15, 2021

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