# If $\left( {2,0} \right),\left( {0,1} \right),\left( {4,5} \right)$ and $\left( {0,c} \right)$ are concyclic then find $c$.

Last updated date: 27th Mar 2023

•

Total views: 309k

•

Views today: 5.86k

Answer

Verified

309k+ views

Hint: Use the general form of circle ${x^2} + {y^2} + 2gx + 2fy + d = 0$.

As we know,

The general equation of circle is ${x^2} + {y^2} + 2gx + 2fy + d = 0..........\left( 1 \right)$

According to question all points $\left( {2,0} \right),\left( {0,1} \right),\left( {4,5} \right)$ and $\left( {0,c} \right)$ are concyclic that means all points satisfy above general equation.

So, we put all points one by one and find the value of $g,f$ and $d$.

Now, put $\left( {2,0} \right)$ in (1) equation

$

4 + 0 + 4g + 0 + d = 0 \\

\Rightarrow 4 + 4g + d = 0...........\left( 2 \right) \\

$

Put $\left( {0,1} \right)$ in (1) equation

$

0 + 1 + 0 + 2f + d = 0 \\

\Rightarrow 1 + 2f + d = 0...........(3) \\

$

Put $(4,5)$ in (1) equation

$

16 + 25 + 8g + 10f + d = 0 \\

\Rightarrow 41 + 8g + 10f + d = 0...........\left( 4 \right) \\

$

Now, we have three equations of three variables.

So, we can easily get $g,f$ and $d$.

Solve $\left( 4 \right) - \left( 2 \right)$ and $\left( 4 \right) - \left( 3 \right)$ equation

$37 + 4g + 10f = 0..........\left( 5 \right)$ and

$

40 + 8g + 8f = 0 \\

\Rightarrow 5 + g + f = 0............\left( 6 \right) \\

\\

$

After solve (5) and (6) equation we get

$f = \frac{{ - 17}}{6}$ and $g = \frac{{ - 13}}{6}$

Now, put these values in (3) equation

$

\Rightarrow 1 - \frac{{34}}{6} + d = 0 \\

\Rightarrow d = \frac{{14}}{3} \\

$

So, equation of circle is ${x^2} + {y^2} - \frac{{13}}{3}x - \frac{{17}}{3}y + \frac{{14}}{3} = 0$

As we know $\left( {0,c} \right)$ also satisfy the equation of circle

$

\Rightarrow 0 + {c^2} - 0 - \frac{{17}}{3}c + \frac{{14}}{3} = 0 \\

\Rightarrow 3{c^2} - 17c + 14 = 0 \Rightarrow 3{c^2} - 3c - 14c + 14 = 0 \\

\Rightarrow 3c\left( {c - 1} \right) - 14\left( {c - 1} \right) = 0 \\

\Rightarrow \left( {c - 1} \right)\left( {3c - 14} \right) = 0 \\

\Rightarrow c = 1,\frac{{14}}{3} \\

$

So, the value of $c$ is $1,\frac{{14}}{3}$ .

Note: Whenever we face such types of problems we use some important points. Like we always use the general equation of circle and remember concyclic points always satisfy the general equation of circle, then after solving some equations we can easily get the required answer.

As we know,

The general equation of circle is ${x^2} + {y^2} + 2gx + 2fy + d = 0..........\left( 1 \right)$

According to question all points $\left( {2,0} \right),\left( {0,1} \right),\left( {4,5} \right)$ and $\left( {0,c} \right)$ are concyclic that means all points satisfy above general equation.

So, we put all points one by one and find the value of $g,f$ and $d$.

Now, put $\left( {2,0} \right)$ in (1) equation

$

4 + 0 + 4g + 0 + d = 0 \\

\Rightarrow 4 + 4g + d = 0...........\left( 2 \right) \\

$

Put $\left( {0,1} \right)$ in (1) equation

$

0 + 1 + 0 + 2f + d = 0 \\

\Rightarrow 1 + 2f + d = 0...........(3) \\

$

Put $(4,5)$ in (1) equation

$

16 + 25 + 8g + 10f + d = 0 \\

\Rightarrow 41 + 8g + 10f + d = 0...........\left( 4 \right) \\

$

Now, we have three equations of three variables.

So, we can easily get $g,f$ and $d$.

Solve $\left( 4 \right) - \left( 2 \right)$ and $\left( 4 \right) - \left( 3 \right)$ equation

$37 + 4g + 10f = 0..........\left( 5 \right)$ and

$

40 + 8g + 8f = 0 \\

\Rightarrow 5 + g + f = 0............\left( 6 \right) \\

\\

$

After solve (5) and (6) equation we get

$f = \frac{{ - 17}}{6}$ and $g = \frac{{ - 13}}{6}$

Now, put these values in (3) equation

$

\Rightarrow 1 - \frac{{34}}{6} + d = 0 \\

\Rightarrow d = \frac{{14}}{3} \\

$

So, equation of circle is ${x^2} + {y^2} - \frac{{13}}{3}x - \frac{{17}}{3}y + \frac{{14}}{3} = 0$

As we know $\left( {0,c} \right)$ also satisfy the equation of circle

$

\Rightarrow 0 + {c^2} - 0 - \frac{{17}}{3}c + \frac{{14}}{3} = 0 \\

\Rightarrow 3{c^2} - 17c + 14 = 0 \Rightarrow 3{c^2} - 3c - 14c + 14 = 0 \\

\Rightarrow 3c\left( {c - 1} \right) - 14\left( {c - 1} \right) = 0 \\

\Rightarrow \left( {c - 1} \right)\left( {3c - 14} \right) = 0 \\

\Rightarrow c = 1,\frac{{14}}{3} \\

$

So, the value of $c$ is $1,\frac{{14}}{3}$ .

Note: Whenever we face such types of problems we use some important points. Like we always use the general equation of circle and remember concyclic points always satisfy the general equation of circle, then after solving some equations we can easily get the required answer.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE