Question

A system of circles is said to be coaxial when every pair of circles has the same radical axis. For coaxial circles, we note that
(1) The centers of all coaxial circles lie in a straight line, which is ⊥ to the common radical axis
(2) Circles passing through two fixed points form a coaxial system with a line joining the points as a common radical axis.
(3) The equation to a coaxial system whose two members are ${{\text{S}}_{\text{1}}}$ and ${{\text{S}}_{\text{2}}}$ is given by ${{\text{S}}_{\text{1}}}$${\text{ + }}\lambda {{\text{S}}_{\text{2}}}$ =0, λ is parameter.
If we take line of centers as x -axis & common radical axis as y-axis, then the simplest form of equation of coaxial circles is given by ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + 2gx + c = 0}}$ where g is variable & c is constant.
If g=±$\sqrt c $ and ${g^2} - c$ vanishes & the circle becomes a point circle. The points (± $\sqrt c $ ,0) are called the limiting points of the system of coaxial circle given by ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + 2gx + c = 0}}$.
On the basis of above information answer the following question:
If (2,1) is a limit point of a coaxial system of circles containing ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ - 6x - 4y - 3 = 0}}$. Then other limit point is
A. (2, 4)
B. (-5, -6)
C. (5, 6)
D. (-2, 4)

Answer 1

Hint: To solve this question we use the basic theory which is mentioned above for any two circles. For better understanding if we summarize above theory then we will be concluded that if we have two circles ${{\text{S}}_1}{\text{ = 0}}$ and ${{\text{S}}_2}{\text{ = 0}}$. And these centers as x -axis & common radical axis as y-axis, then the simplest form of equation of coaxial circles is given by ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + 2gx + c = 0}}$ where g is variable & c is constant. And also The equation to a coaxial system whose two members are ${{\text{S}}_{\text{1}}}$ and ${{\text{S}}_{\text{2}}}$ is given by ${{\text{S}}_{\text{1}}}$${\text{ + }}\lambda {{\text{S}}_{\text{2}}}$ =0, λ is parameter.

Complete step by step solution:
Given, one limiting point is (2,1)
∴ Point circle with the help of limiting point (2,1) is given by
${\left( {x - 2} \right)^2} + {\left( {y - 1} \right)^2} = 0$& other member of the family is ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ - 6x - 4y - 3 = 0}}$.
So, family of coaxial circle is given by ${{\text{S}}_{\text{1}}}$${\text{ + }}\lambda {{\text{S}}_{\text{2}}}$ =0
$ \Rightarrow $${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ - 6x - 4y - 3 + }}\lambda ({{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ - 4x - 2y + 5) = 0}}$
$ \Rightarrow $$\left( {1 + \lambda } \right)\left( {{x^2} + {y^2}} \right) - 2x\left( {3 + 2\lambda } \right) - 2y\left( {2 + \lambda } \right) + \left( { - 3 + 5\lambda } \right) = 0$
$ \Rightarrow $Center is
$ \Rightarrow $$\left( {\dfrac{{3 + 2\lambda }}{{1 + \lambda }},\dfrac{{2 + \lambda }}{{1 + \lambda }}} \right)$ ……………. (1)
$\therefore $ Radius = $\sqrt {{{\left( {3 + 2\lambda } \right)}^2} + {{\left( {2 + \lambda } \right)}^2} - \left( { - 3 + 5\lambda } \right)\left( {1 + \lambda } \right)} = 0$
$ \Rightarrow $$\left( {5{\lambda ^2} + 16\lambda + 13} \right) - \left( {5{\lambda ^2} + 2\lambda - 3} \right) = 0$
$ \Rightarrow $$14\lambda = - 16$
$ \Rightarrow $$\lambda = \dfrac{{ - 8}}{7}$
So, we get limiting point by using (1)
∴ Other limiting point is (−5,−6).
Therefore, option (B) is the correct answer.

Note: A set of circles is coaxial if every pair of circles from the system have the same radical axis. Thus, a coaxial system of circles is defined by the radical axis and any one of the circles. And also We need to remember some basic properties of circles. For example, the outer line of a circle is at equidistant from the center. The diameter of the circle divides it into two equal parts. Circles which have equal radii are congruent to each other.