Question

A: The radical center of the circles ${x^2} + {y^2} = 4,\;{x^2} + {y^2} - 3x = 4,\;{x^2} + {y^2} - 4y = 4$ is (0, 0)
R: Radical center of three circles is the point of concurrence of the radical axes of the circles taken in pairs.
A) Both A and R are true and R is the correct explanation of A
B) Both A and R are true and R is not the correct explanation of A
C) A is true but R is false
D) A is false but R is true

Answer 1

Hint: Radical Axis Formula:-
Let two circles be
\[S:{x^2} + {y^2} + 2gx + 2fy + c = 0\;\] and
\[S\prime :{x_2} + {y_2} + 2g\prime x + 2f\prime y + c\prime = 0\]
then equation of radical axis is
 \[S - S\prime = 0\]
i.e. \[\;2(g - g\prime )x + 2(f - f\prime ) + (c - c\prime ) = 0\]
Radical Centre definition: -
The radical axis of three circles, taken in pairs, meet at a point, which is called their radical centre. The coordinate of radical centre can be found by solving the equation
\[{S_1} = {S_2} = {S_3} = 0\]

Complete step-by-step answer:
Let:
Radical axis of first circle\[{S_1} = {x^2} + {y^2} = 4............................(1)\]
Radical axis of second circle \[{S_2} = {x^2} + {y^2} - 3x = 4....................(2)\]
Radical axis of third circle \[{S_3} = {x^2} + {y^2} - 4y = 4....................(3)\]
Radical axis of two circles is the locus of a point, which moves such that length of tangent drawn from it to the two circles are equal.
Equation of radical axis is${S_1} - {S_2} = 0$. Hence,
$
  {S_1} - {S_2} = 0;\; \\
  {S_2} - {S_3} = 0;\; \\
  {S_3} - {S_1} = 0 \\
 $
We will get the equation of radical axis and by solving them we can obtain the required radical center. Radical axis is perpendicular to the straight line joining the centre of circle. So, two circles are non-concentric.
Radical axis of first and second circle is given by i.e. Subtracting equation (1) and (2)
\[
   \Rightarrow \left( {{x^2} + {y^2}} \right) - \left( {{x^2} + {y^2} - 3x} \right) = 0 \\
   \Rightarrow 3x = 0 \\
   \Rightarrow x = 0..............................(4) \\
 \]
Radical axis of second and third circle is given by i.e. Subtracting equation (2) and (3)
$
   \Rightarrow \left( {{x^2} + {y^2} - 3x} \right) - \left( {{x^2} + {y^2} - 4y} \right) = 0 \\
   \Rightarrow 3x + 4y = 0..............................(5) \\
 $
Radical axis of third and second circle is given by i.e. Subtracting equation (3) and (1)
$
   \Rightarrow \left( {{x^2} + {y^2} - 4y} \right) - \left( {{x^2} + {y^2}} \right) = 0 \\
   \Rightarrow - 4y = 0 \\
   \Rightarrow y = 0..............................(6) \\
 $
 Thus the radical centre is (0, 0). Hence A is correct.
 Substituting x in equation (5) by equation (4) gives the value of y similar to given by equation (6). These three straight lines meet at a point; since it is obvious that the values of x and y which simultaneously satisfy two of the equations, is also satisfying the third equation.
Hence R is also correct.
So Option (A) is the correct answer.

Note: Radical axis is a useful tool to prove the collinearity of points. To prove that a certain number of points lie on a straight line, we may try to show that each of these points has equal power to two circles and then draw the conclusion that they all lie on the radical axis of the circles.
Similarly, we can use radical center to prove the concurrency of lines. The strategy is to show that each of these lines is the radical axis of a pair of circles and then conclude that they meet at the radical center.