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# Find the center of the circle passing through the points \$(-1,0)\$, \$(1,0)\$, and \$(3,1)\$. Express your answer in the form '\$(a,b)\$.'

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Find the center of the circle passing through the points \$(-1,0)\$, \$(1,0)\$, and \$(3,1)\$. Express your answer in the form "\$(a,b)\$."

michaelcai  Oct 30, 2017

#1
+6312
+2

The distance from the center of the circle to each of the  3  points will be the same.

We can call this distance  " r "  for radius.  So we know...

(-1 - a)2 + (0 - b)2  =  r2

(1 - a)2 + (0 - b)2  =  r2            And...

(3 - a)2 + (1 - b)2  =  r2

Set the first two values of  r2  equal to each other.

(-1 - a)2  + b2  =  (1 - a)2 + b2

Subtract  b2  from both sides.

(-1 - a)2  =  (1 - a)2

Take the  ±  square root of both sides.

-1 - a  =  ±(1 - a)

-1 - a  =  1 - a     or     -1 - a  =  -1 + a     →     a  =  0

Using this value for  a , set the second and third values of  r2  equal to each other.

(1 - 0)2 + b2  =  32 + (1 - b)2           Solve this for  b .

1 + b2  =  9 + 1 - 2b + b2

1  =  10 - 2b

-9  =  -2b

9/2  =  b

So the center of the circle is  (0, 9/2) .  Here's a graph.

hectictar  Oct 31, 2017
Sort:

#1
+6312
+2

The distance from the center of the circle to each of the  3  points will be the same.

We can call this distance  " r "  for radius.  So we know...

(-1 - a)2 + (0 - b)2  =  r2

(1 - a)2 + (0 - b)2  =  r2            And...

(3 - a)2 + (1 - b)2  =  r2

Set the first two values of  r2  equal to each other.

(-1 - a)2  + b2  =  (1 - a)2 + b2

Subtract  b2  from both sides.

(-1 - a)2  =  (1 - a)2

Take the  ±  square root of both sides.

-1 - a  =  ±(1 - a)

-1 - a  =  1 - a     or     -1 - a  =  -1 + a     →     a  =  0

Using this value for  a , set the second and third values of  r2  equal to each other.

(1 - 0)2 + b2  =  32 + (1 - b)2           Solve this for  b .

1 + b2  =  9 + 1 - 2b + b2

1  =  10 - 2b

-9  =  -2b

9/2  =  b

So the center of the circle is  (0, 9/2) .  Here's a graph.

hectictar  Oct 31, 2017

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