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A circle is tangent to the $y$-axis at the point $(0,2)$ and passes through the point $(8,0)$, as shown. Find the radius of the circle.

michaelcai  Nov 6, 2017

Best Answer 

 #2
avatar+5261 
+1

The circle is tangent to the y-axis at  (0, 2) , so the y-coordinate of the center must be  2  .

 

We know..

 

(x - h)2 + (y - 2)2  =  r2      , where  h  is the x-coordinate of the vertex and  r  is the radius.

 

(0 - h)2 + (2 - 2)2  =  r2     →​     h2  =  r2     →     h  =  ± r

 

(8 - h)2 + (0 - 2)2  =  r2     →​     (8 - h)2 + 4  =  r2

 

In the last equation, substitute  ± r  in for  h .

 

(8 - ± r)2 + 4  =  r2

 

(8 + r)2 + 4  =  r2          or          (8 - r)2 + 4  =  r2
64 + 16r + r2 + 4  =  r2 64 - 16r + r2 + 4  =  r2
68 + 16r  =  0 68 - 16r  =  0
r  =  -68/16 r  =  68/16
r  =  -4.25 r  =  4.25

 

Since  r  is a distance,  r  =  4.25

 

Here is a graph: https://www.desmos.com/calculator/hipxsimqkb    smiley

hectictar  Nov 6, 2017
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3+0 Answers

 #1
avatar+78762 
0

We need to see a diagram or -  at least -  know another point

 

 

cool cool cool

CPhill  Nov 6, 2017
 #2
avatar+5261 
+1
Best Answer

The circle is tangent to the y-axis at  (0, 2) , so the y-coordinate of the center must be  2  .

 

We know..

 

(x - h)2 + (y - 2)2  =  r2      , where  h  is the x-coordinate of the vertex and  r  is the radius.

 

(0 - h)2 + (2 - 2)2  =  r2     →​     h2  =  r2     →     h  =  ± r

 

(8 - h)2 + (0 - 2)2  =  r2     →​     (8 - h)2 + 4  =  r2

 

In the last equation, substitute  ± r  in for  h .

 

(8 - ± r)2 + 4  =  r2

 

(8 + r)2 + 4  =  r2          or          (8 - r)2 + 4  =  r2
64 + 16r + r2 + 4  =  r2 64 - 16r + r2 + 4  =  r2
68 + 16r  =  0 68 - 16r  =  0
r  =  -68/16 r  =  68/16
r  =  -4.25 r  =  4.25

 

Since  r  is a distance,  r  =  4.25

 

Here is a graph: https://www.desmos.com/calculator/hipxsimqkb    smiley

hectictar  Nov 6, 2017
 #3
avatar+78762 
+1

Ah....thanks, hectictar.....I forgot about the word "tangent"    !!!!!

 

 

cool cool cool

CPhill  Nov 7, 2017

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