
The four distinct points $(0,0),(2,0),(0, - 2)$ and $(k, - 2)$ are con-cyclic. Then find the value of k.
A. 2
B. -2
C. 0
D. 1
Answer
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Hint: Assume ${x^2} + {y^2} + 2gx + 2fy + c = 0$ as the equation of a cycle. Now, select three points $(0,0),(2,0),(0, - 2)$from the given four points in the question and substitute them in the equation of the circle to form three equations containing the variables g, f and c. Solve the equations to find the value of g, f and c. Finally, substitute the fourth point $(k, - 2)$in the equation of the circle and obtain the value of k.
Complete step by step solution:
The given points are $(0,0),(2,0),(0, - 2)$ and $(k, - 2)$.
Now, substitute the points $(0,0),(2,0),(0, - 2)$one by one in the equation of the cycle${x^2} + {y^2} + 2gx + 2fy + c = 0$ to obtain the values of g, f and c.
Substitute (0,0) for (x, y), then
${0^2} + {0^2} + 2g.0 + 2f.0 + c = 0$
$ \Rightarrow c = 0$
Now, Substitute (2,0) for (x, y) in ${x^2} + {y^2} + 2gx + 2fy = 0$, as c=0, then
${2^2} + {0^2} + 2g.2 + 2f.0 = 0$
$ \Rightarrow 4 + 4g = 0$
$ \Rightarrow g = - 1$
Now, Substitute (0, -2) for (x, y) in ${x^2} + {y^2} - 2x + 2fy = 0$, as c=0 and g=-1, then
${0^2} + {( - 2)^2} - 2.0 + 2f.( - 2) = 0$
$ \Rightarrow 4 - 4f = 0$
$ \Rightarrow f = 1$
Therefore, the cycle equation is ${x^2} + {y^2} - 2x + 2y = 0$.
So, substitute (k, -2) in the equation ${x^2} + {y^2} - 2x + 2y = 0$ to obtain the value of k.
${k^2} + {( - 2)^2} - 2k + 2( - 2) = 0$
${k^2} + 4 - 2k - 4 = 0$
${k^2} - 2k = 0$
$k(k - 2) = 0$
Hence, k=0 or 2.
Option ‘A or C’ is correct
Note: One may note that we have chosen the order of substituting the points as$(0,0),(2,0),(0, - 2)$, but there is no particular rule that we have to choose the points in this order, the order of the points is not important here. In any order we will get the values of f, g and c as 1, -1 and 0 respectively.
Complete step by step solution:
The given points are $(0,0),(2,0),(0, - 2)$ and $(k, - 2)$.
Now, substitute the points $(0,0),(2,0),(0, - 2)$one by one in the equation of the cycle${x^2} + {y^2} + 2gx + 2fy + c = 0$ to obtain the values of g, f and c.
Substitute (0,0) for (x, y), then
${0^2} + {0^2} + 2g.0 + 2f.0 + c = 0$
$ \Rightarrow c = 0$
Now, Substitute (2,0) for (x, y) in ${x^2} + {y^2} + 2gx + 2fy = 0$, as c=0, then
${2^2} + {0^2} + 2g.2 + 2f.0 = 0$
$ \Rightarrow 4 + 4g = 0$
$ \Rightarrow g = - 1$
Now, Substitute (0, -2) for (x, y) in ${x^2} + {y^2} - 2x + 2fy = 0$, as c=0 and g=-1, then
${0^2} + {( - 2)^2} - 2.0 + 2f.( - 2) = 0$
$ \Rightarrow 4 - 4f = 0$
$ \Rightarrow f = 1$
Therefore, the cycle equation is ${x^2} + {y^2} - 2x + 2y = 0$.
So, substitute (k, -2) in the equation ${x^2} + {y^2} - 2x + 2y = 0$ to obtain the value of k.
${k^2} + {( - 2)^2} - 2k + 2( - 2) = 0$
${k^2} + 4 - 2k - 4 = 0$
${k^2} - 2k = 0$
$k(k - 2) = 0$
Hence, k=0 or 2.
Option ‘A or C’ is correct
Note: One may note that we have chosen the order of substituting the points as$(0,0),(2,0),(0, - 2)$, but there is no particular rule that we have to choose the points in this order, the order of the points is not important here. In any order we will get the values of f, g and c as 1, -1 and 0 respectively.
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