Question

Show that the four points (-6, 0), (-2, 2), (-2, -8) and (1, 1) are concyclic.

Answer 1

Hint: Assume \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c\] as the equation of a cycle. Now, select any three points 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 in the equation of the circle, if the point satisfies the equation of the circle then the four points will be concyclic otherwise not.

Complete step-by-step answer:
Here, we have been provided with four points (-6, 0), (-2, 2), (-2, -8) and (1, 1) and we have to show that they are concyclic. First let us see what are concyclic points.
In geometry, a set of points are said to be concyclic if they lie on a common circle. All concyclic points are at the same distance from the centre of the circle. So, the given four points in the question will be concyclic if they all lie on the same circle.
Let us assume a circle whose equation is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] and points (-6, 0), (-2, 2) and (-2, -8) lie on it. Therefore, these points will satisfy the equation of the circle, so we have,
(1) Substituting (-6, 0) in the equation of the circle, we get,
\[\Rightarrow 36+0-12g+0+c=0\]
\[\Rightarrow 36-12g+c=0\] - (1)
(2) Substituting (-2, 2) in the equation of the circle, we get,
\[\Rightarrow 4+4-4g+4f+c=0\]
\[\Rightarrow 8-4g+4f+c=0\] - (ii)
(3) Substituting (-2, -8) in the equation of the circle, we get,
\[\Rightarrow 4+64-4g-16f+c=0\]
\[\Rightarrow 68-4g-16f+c=0\] - (iii)
Now, solving the above three equations, we get,
\[\Rightarrow g=2,f=3\] and c = -12
Therefore, the equation of the circle becomes: -
\[\Rightarrow {{x}^{2}}+{{y}^{2}}+4x+6y-12=0\]
Now, the fourth point (1, 1) will be concyclic if and only if it satisfies the above obtained equation of the circle. So, substituting this point in the equation of the circle, we get,
\[\Rightarrow 1+1+4+6-12=0\]
\[\Rightarrow \] L.H.S = R.H.S
Hence, the point (1, 1) satisfies the given equation of the circle. Therefore, the given points are concyclic, i.e. they lie on a common circle.
Hence proved.

Note: One may note that we have chosen (-6, 0), (-2, 2) and (-2, 8) to find the values of g, f and c from the equation of the circle. You may choose any other three points from the four points given. The values of g, f and c will remain the same. Remember that if we will join all the points to form a quadrilateral then it will be known as a cyclic quadrilateral.