Question

For value of $k$, the points $\left( {0,0} \right),\left( {1,3} \right),\left( {2,4} \right)$, and $\left( {k,3} \right)$ are concyclic?
A. 4
B. 9
C. 6
D. 8

Answer 1

Hint: First, substitute the given points $\left( {0,0} \right),\left( {1,3} \right)$, and $\left( {2,4} \right)$ in the standard equation of a circle to calculate the values of $c,f$, and $g$. Then, simplify the standard equation of a circle by substituting the values. And in the end, substitute the point $\left( {k,3} \right)$ in the simplified equation of a circle and solve it to reach the required answer.

Formula Used:
The standard equation of a circle is: ${x^2} + {y^2} + 2gx + 2fy + c = 0$, where $\left( { - g, - f} \right)$ is the center of a circle.

Complete step by step solution:
Given: The points $\left( {0,0} \right),\left( {1,3} \right),\left( {2,4} \right)$, and $\left( {k,3} \right)$ are concyclic.
We know that if points are concyclic, then they lie on the same circle.
Substitute the point $\left( {0,0} \right)$ in the standard equation of a circle.
We get,
${0^2} + {0^2} + 2g\left( 0 \right) + 2f\left( 0 \right) + c = 0$
$ \Rightarrow c = 0$ $.....\left( 1 \right)$

Now substitute the point $\left( {1,3} \right)$ in the standard equation of a circle.
We get,
${1^2} + {3^2} + 2g\left( 1 \right) + 2f\left( 3 \right) = 0$
$ \Rightarrow 1 + 9 + 2g + 6f = 0$
$ \Rightarrow 10 + 2g + 6f = 0$ $.....\left( 2 \right)$

Now substitute the point $\left( {2,4} \right)$ in the standard equation of a circle.
We get,
${2^2} + {4^2} + 2g\left( 2 \right) + 2f\left( 4 \right) = 0$
$ \Rightarrow 4 + 16 + 4g + 8f = 0$
$ \Rightarrow 20 + 4g + 8f = 0$
Divide both sides by 2.
$10 + 2g + 4f = 0$
$ \Rightarrow 10 + 2g = - 4f$ $.....\left( 3 \right)$

Substitute the equation $\left( 3 \right)$ in the equation $\left( 2 \right)$.
$ - 4f + 6f = 0$
$ \Rightarrow 2f = 0$
$ \Rightarrow f = 0$ $.....\left( 4 \right)$
Substitute the equation $\left( 4 \right)$ in the equation $\left( 2 \right)$.
$10 + 2g + 6\left( 0 \right) = 0$
$ \Rightarrow 2g = - 10$
Divide both sides by 2.
$ \Rightarrow g = - 5$ $.....\left( 5 \right)$

Now substitute the equations $\left( 1 \right)$, $\left( 4 \right)$ and $\left( 5 \right)$ in the standard equation of a circle.
${x^2} + {y^2} + 2\left( { - 5} \right)x + 2\left( 0 \right)y + 0 = 0$
$ \Rightarrow {x^2} + {y^2} - 10x = 0$
Substitute the point $\left( {k,3} \right)$ in the above equation.
${k^2} + {3^2} - 10k = 0$
$ \Rightarrow {k^2} - 10k + 9 = 0$
Factorize the quadratic equation.
${k^2} - 9k - k + 9 = 0$
$ \Rightarrow k\left( {k - 9} \right) - \left( {k - 9} \right) = 0$
$ \Rightarrow \left( {k - 9} \right)\left( {k - 1} \right) = 0$
$ \Rightarrow k - 9 = 0$ or $k - 1 = 0$
$ \Rightarrow k = 9$ or $k = 1$
Thus, the values of $k$ are: $k = 1$ or $k = 9$

Option ‘B’ is correct

Note: Concyclic points are defined as any two or more points that lie on the common circle. All concyclic points are the same distance from the center of the circle.