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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.

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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.

Nov 6, 2017

#2
+7350
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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

Nov 6, 2017

#1
+98044
0

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

Nov 6, 2017
#2
+7350
+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

hectictar Nov 6, 2017
#3
+98044
+1

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

Nov 7, 2017