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Question: If \(\left( {2,0} \right),\left( {0,1} \right),\left( {4,5} \right)\) and \(\left( {0,c} \right)\) a...

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

Explanation

Solution

Hint: Use the general form of circle x2+y2+2gx+2fy+d=0{x^2} + {y^2} + 2gx + 2fy + d = 0.

As we know,
The general equation of circle is x2+y2+2gx+2fy+d=0..........(1){x^2} + {y^2} + 2gx + 2fy + d = 0..........\left( 1 \right)
According to question all points (2,0),(0,1),(4,5)\left( {2,0} \right),\left( {0,1} \right),\left( {4,5} \right) and (0,c)\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,fg,f and dd.
Now, put (2,0)\left( {2,0} \right) in (1) equation
4+0+4g+0+d=0 ⇒4+4g+d=0...........(2)  4 + 0 + 4g + 0 + d = 0 \\\ \Rightarrow 4 + 4g + d = 0...........\left( 2 \right) \\\
Put (0,1)\left( {0,1} \right) in (1) equation
0+1+0+2f+d=0 ⇒1+2f+d=0...........(3)  0 + 1 + 0 + 2f + d = 0 \\\ \Rightarrow 1 + 2f + d = 0...........(3) \\\
Put (4,5)(4,5) in (1) equation
16+25+8g+10f+d=0 ⇒41+8g+10f+d=0...........(4)  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,fg,f and dd.
Solve (4)−(2)\left( 4 \right) - \left( 2 \right) and (4)−(3)\left( 4 \right) - \left( 3 \right) equation
37+4g+10f=0..........(5)37 + 4g + 10f = 0..........\left( 5 \right) and
40+8g+8f=0 ⇒5+g+f=0............(6)   40 + 8g + 8f = 0 \\\ \Rightarrow 5 + g + f = 0............\left( 6 \right) \\\ \\\
After solve (5) and (6) equation we get
f=−176f = \frac{{ - 17}}{6} and g=−136g = \frac{{ - 13}}{6}
Now, put these values in (3) equation
⇒1−346+d=0 ⇒d=143  \Rightarrow 1 - \frac{{34}}{6} + d = 0 \\\ \Rightarrow d = \frac{{14}}{3} \\\
So, equation of circle is x2+y2−133x−173y+143=0{x^2} + {y^2} - \frac{{13}}{3}x - \frac{{17}}{3}y + \frac{{14}}{3} = 0
As we know (0,c)\left( {0,c} \right) also satisfy the equation of circle
⇒0+c2−0−173c+143=0 ⇒3c2−17c+14=0⇒3c2−3c−14c+14=0 ⇒3c(c−1)−14(c−1)=0 ⇒(c−1)(3c−14)=0 ⇒c=1,143  \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 cc is 1,1431,\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.