Question

Find the equation of the circle passing through the points $P(3,4),Q(3, - 6) and{\text{ }}R( - 1,2)$

Answer 1

Hint-It can be solved by starting with the general form of the circle. After substitution of given points in the general equation of the circle, we will get three equations. So we will have three unknown variables and three equations which can be solved easily.
Complete step-by-step answer:
Let the general equation of circle is
${x^2} + {y^2} + Dx + Ey + F = 0$
Now this circle passes through three points as mentioned in the question.
So, when this circle passes through point (3,4), the equation we get is
$ \Rightarrow 9 + 16 + 3D + 4E + F = 0$
$ \Rightarrow 25 + 3D + 4E + F = 0$ -----(1)
Similarly, when this circle passes through point (3,-6), the equation we get is
$ \Rightarrow 9 + 36 + 3D - 6E + F = 0$
$ \Rightarrow 45 + 3D - 6E + F = 0$ ------(2)
Similarly, when this circle passes through point (-1,2), the equation we get is
$ \Rightarrow 1 + 4 - D + 2E + F = 0$
$ \Rightarrow 5 - D + 2E + F = 0$ -------(3)
Now, we have three unknown variables D, E, F and three equations.
Subtracting equation 2 from 1, we get
$
   \Rightarrow - 20 + 10E = 0 \\
   \Rightarrow 10E = 20 \\
   \Rightarrow E = 2 \\
$
Now subtracting equation 3 from equation 2, we get
$
   \Rightarrow 40 + 4D - 8E = 0 \\
  Now,{\text{assigning the value of E}} \\
   \Rightarrow 40 + 4D - 16 = 0 \\
   \Rightarrow 4D = - 24 \\
   \Rightarrow D = - 8 \\
$
Now, putting the value of D, E in any equation. So let us assign these values in equation 3, we get
$
   \Rightarrow 5 - D + 2E + F = 0 \\
   \Rightarrow 5 + 8 + 4 + F = 0 \\
   \Rightarrow F = - 17 \\
$
$
  \therefore {\text{The equation of the circle becomes}} \\
   \Rightarrow {{\text{x}}^2} + {y^2} - 8x + 2y - 17 = 0 \\
   \Rightarrow ({x^2} - 8x) + ({y^2} + 2y) = 17 \\
   \Rightarrow {(x - 4)^2} + {(y + 1)^2} = 34 \\
$
Which when compared with the standard equation gives the Centre (4,-1) and radius $\sqrt {34} $
Note-These types of questions can be solved also by using the perpendicular bisector of two chords. We need to find the midpoints of two chords with analysing their gradient. Then solving the equations of perpendicular bisectors using gradient point form for the centre and radius. Then use the centre-radius form to get the equation of the circle.