Question

Find the value of c if the points $\left( {2,0} \right),\left( {0,1} \right),\left( {4,5} \right){\text{ and }}\left( {0,c} \right)$ are concyclic.

Answer 1

Hint: in order to solve the problem use the property of concyclic points. Consider the general equation for the circle and find out the constants with the help of given 3 points other than one to solve. Finally put the values of the last coordinate to find the unknown value.

Complete Step-by-Step solution:
Given points are: $\left( {2,0} \right),\left( {0,1} \right),\left( {4,5} \right){\text{ and }}\left( {0,c} \right)$
Let us name the given points as \[A\left( {2,0} \right),B\left( {0,1} \right),C\left( {4,5} \right){\text{ and }}D\left( {0,c} \right)\]
We know the general equation for the center \[\left( { - g, - f} \right)\] is ${x^2} + {y^2} + 2gx + 2fy + h = 0$
Now let us the values from the coordinates given:
First let us put the values of point D in the general equation.
\[
  \because {x^2} + {y^2} + 2gx + 2fy + h = 0{\text{ for }}D\left( {0,c} \right) \\
   \Rightarrow {\left( 0 \right)^2} + {\left( c \right)^2} + 2g\left( 0 \right) + 2f\left( c \right) + h = 0 \\
   \Rightarrow {c^2} + 2cf + h = 0..............(1) \\
 \]
From the above equation, in order to find the value of c, we need the values of h and f.
Now by taking the point A for the same circle we get:
\[
  \because {x^2} + {y^2} + 2gx + 2fy + h = 0{\text{ for }}A\left( {2,0} \right) \\
   \Rightarrow {\left( 2 \right)^2} + {\left( 0 \right)^2} + 2g\left( 2 \right) + 2f\left( 0 \right) + h = 0 \\
   \Rightarrow 4 + 4g + h = 0...........(2) \\
 \]
Now by taking the point B for the same circle we get:
\[
  \because {x^2} + {y^2} + 2gx + 2fy + h = 0{\text{ for }}B\left( {0,1} \right) \\
   \Rightarrow {\left( 0 \right)^2} + {\left( 1 \right)^2} + 2g\left( 0 \right) + 2f\left( 1 \right) + h = 0 \\
   \Rightarrow 1 + 2f + h = 0...........(3) \\
 \]
Now by taking the point C for the same circle we get:
\[
  \because {x^2} + {y^2} + 2gx + 2fy + h = 0{\text{ for }}C\left( {4,5} \right) \\
   \Rightarrow {\left( 4 \right)^2} + {\left( 5 \right)^2} + 2g\left( 4 \right) + 2f\left( 5 \right) + h = 0 \\
   \Rightarrow 16 + 25 + 8g + 10f + h = 0 \\
   \Rightarrow 41 + 8g + 10f + h = 0...........(4) \\
 \]
Now from the above three equations, we need to find out the constant terms.
Subtracting equation (3) from equation (2) we can eliminate h
$
   \Rightarrow \left( {4 + 4g + h} \right) - \left( {1 + 2f + h} \right) = 0 - 0 \\
   \Rightarrow 3 + 4g - 2f = 0 \\
 $
In order to subtract the above equation from equation (4) we need to bring the coefficients equal. So we multiply the above equation by 2.
$
   \Rightarrow 2 \times \left( {3 + 4g - 2f} \right) = 2 \times 0 \\
   \Rightarrow 6 + 8g - 4f = 0............(5) \\
 $
To proceed further we will subtract equation (5) from equation (4)
\[
   \Rightarrow \left( {6 + 8g - 4f} \right) - \left( {41 + 8g + 10f + h} \right) = 0 - 0 \\
   \Rightarrow - 35 - 14f - h = 0..............(6) \\
 \]
Now we have two equations. Equation (6) and equation (3) with just two variables f and h
Solving these two equations
Adding equation (6) and equation (3) we get:
$
   \Rightarrow 1 + 2f + h + \left( { - 35 - 14f - h} \right) = 0 - 0 \\
   \Rightarrow - 12f - 34 = 0 \\
   \Rightarrow f = \dfrac{{34}}{{ - 12}} = \dfrac{{ - 17}}{6} \\
$
Now, substituting the value of f in equation (3), we get
$
  1 + 2\left( {\dfrac{{ - 17}}{6}} \right) + h = 0 \\
   \Rightarrow h = \dfrac{{34}}{6} - 1 = \dfrac{{28}}{6} = \dfrac{{14}}{3} \\
 $
Now, putting the values of h and f in equation (1), we get:
\[
   \Rightarrow {c^2} + 2cf + h = 0 \\
   \Rightarrow {c^2} + 2c\left( {\dfrac{{ - 17}}{6}} \right) + \dfrac{{14}}{3} = 0 \\
 \]
Solving the quadratic equation in order to find the value of c
$
   \Rightarrow 3{c^2} - 17c + 14 = 0 \\
   \Rightarrow 3{c^2} - 14c - 3c + 14 = 0 \\
   \Rightarrow 3c\left( {c - 1} \right) - 14\left( {c - 1} \right) = 0 \\
   \Rightarrow \left( {3c - 14} \right)\left( {c - 1} \right) = 0 \\
   \Rightarrow \left( {3c - 14} \right) = 0{\text{ or }}\left( {c - 1} \right) = 0 \\
   \Rightarrow c = \dfrac{{14}}{3}{\text{ or }}c = 1 \\
 $
For different values of c the points are $\left( {0,1} \right){\text{ and }}\left( {0,\dfrac{{14}}{3}} \right)$
Since, the point $\left( {0,1} \right)$ is already mentioned in the problem so the point is $D\left( {0,\dfrac{{14}}{3}} \right)$ .
Hence the final concyclic point is $\left( {0,\dfrac{{14}}{3}} \right)$

Note: A set of points are said to be concyclic if they lie on a common circle. All concyclic points are the same distance from the center of the circle. Three points in the plane that do not all fall on a straight line are concyclic, but four or more such points in the plane are not necessarily concyclic. Students must remember the general formula for different shapes such as the circle which is mentioned in the solution.