Question

If a circle passes through \[\left( {0,0} \right),\left( {a,0} \right)\], and \[\left( {0,b} \right)\], then the coordinates of its center are
A. \[\left( {\dfrac{b}{2},\dfrac{a}{2}} \right)\]
B. \[\left( {\dfrac{a}{2},\dfrac{b}{2}} \right)\]
C. \[\left( {b,a} \right)\]
D. \[\left( {a,b} \right)\]

Answer 1

Hint: In this question, we need to find the coordinates of its center of circle passing through given points. We know the general equation of a circle is \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]. Now, substitute the values of given coordinates then solve two equations to get the desired result.

Formula used:
We have been using the following formulas:
1. \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]

Complete step-by-step solution:
Given coordinates are \[\left( {0,0} \right),\left( {a,0} \right)\] and \[\left( {0,b} \right)\]
Now we know that the equation of circle having a center (h, k) having radius as r units, is
\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}...\left( 1 \right)\]
Now we substitute the values of given coordinates in the above equation (1), and we get
At coordinate \[\left( {0,0} \right)\]:
\[
  {\left( {0 - h} \right)^2} + {\left( {0 - k} \right)^2} = {r^2} \\
  {\left( { - h} \right)^2} + {\left( { - k} \right)^2} = {r^2} \\
  {h^2} + {k^2} = {r^2}...\left( 2 \right)
 \]
Now at coordinate \[\left( {a,0} \right)\]:
\[
  {\left( {a - h} \right)^2} + {\left( {0 - k} \right)^2} = {r^2} \\
  {\left( {a - h} \right)^2} + {\left( { - k} \right)^2} = {r^2}...\left( 3 \right)
 \]
We know that \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
Now by applying the above identity in above equation (3), we get
\[{a^2} + {h^2} - 2ah + {k^2} = {r^2}...\left( 4 \right)\]
Now at coordinate \[\left( {0,b} \right)\]
\[
  {\left( {0 - h} \right)^2} + {\left( {b - k} \right)^2} = {r^2} \\
  {\left( { - h} \right)^2} + {\left( {b - k} \right)^2} = {r^2}...\left( 5 \right)
 \]
We know that \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
Now by applying the above identity in above equation (5), we get
\[{h^2} + {b^2} - 2bk + {k^2} = {r^2}...\left( 6 \right)\]
Now we subtract equation (6) from equation (4), we get
\[{a^2} - {b^2} - 2ah + 2bk = 0\]
On further simplification, we get
\[
  {a^2} - 2ah - {b^2} + 2bk = 0 \\
  a\left( {a - 2h} \right) + b\left( {b - 2k} \right) = 0
 \]
Now break the above terms in two parts:
\[
  a\left( {a - 2h} \right) = 0 \\
  b\left( {b - 2k} \right) = 0
 \]
On simplifying above equation:
\[
  a = 0, \\
  a - 2h = 0 \\
  a = 2h
 \]
And
\[
  b = 0, \\
  b - 2k = 0 \\
  b = 2k
 \]
Now by simplifying the above equation, we get
\[
  a = 2h \\
  h = \dfrac{a}{2}
 \]
And
\[
  b = 2k \\
  k = \dfrac{b}{2}
 \]
Therefore, the coordinates of its center are \[\left( {\dfrac{a}{2},\dfrac{b}{2}} \right)\]
Hence, option (B) is correct

Note: To answer the question above, students must be familiar with the circle equation. The student needs to be careful while entering the coordinates into the equation and when solving the equation after doing so.