Question

Find value of $$k$$ if $$kx + 3y - 1 = 0,2x + y + 5 = 0$$ are conjugate lines with respective to circle $${x^2} + {y^2} - 2x - 4y - 4 = 0$$

Answer 1

Hint: If two lines are conjugate line with respect to circle $x^2+y^2=r^2$ then that satisfies certain equation which is $${r^2}({l_1}{l_2} + {m_1}{m_2}) = {n_1}{n_2}$$. Here we have equate given two lines with $${l_1}x + {m_1}y + {n_1} = 0$$ and $${l_2}x + {m_2}y + {n_2} = 0$$ respectively. After that we also have a circle equation, we have to convert the given circle equation in standard form. Then we have put proper value in equation $${r^2}({l_1}{l_2} + {m_1}{m_2}) = {n_1}{n_2}$$ so that we can equate and can find correct answer.

Complete step-by-step answer:
Let’s consider two lines $${l_1}x + {m_1}y + {n_1} = 0$$ and $${l_2}x + {m_2}y + {n_2} = 0$$ are conjugate line with respect to circle $${x^2} + {y^2} = {r^2}$$ . In this condition it satisfies the equation $${r^2}({l_1}{l_2} + {m_1}{m_2}) = {n_1}{n_2}$$ …….. (1)
We have equations $$kx + 3y - 1 = 0,2x + y + 5 = 0$$ and $${x^2} + {y^2} - 2x - 4y - 4 = 0$$
We can reform the circle equation as $${(x - 1)^2} + {(y - 2)^2} = 9$$ now we have to convert the equation $${(x - 1)^2} + {(y - 2)^2} = 9$$ in $${x^2} + {y^2} = {r^2}$$ form, for that we have to consider $$X = x - 1,Y = y - 2$$ then circle equation will be $${X^2} + {Y^2} = 9$$
Put values of $$x$$ and $$y$$ in given equation $$kx + 3y - 1 = 0,2x + y + 5 = 0$$
Equation will become
 $$\eqalign{
  & k(X + 1) + 3(Y + 2) - 1 = 0 \cr
  & \Rightarrow kX + k + 3Y + 6 - 1 = 0 \cr
  & \Rightarrow kX + 3Y + k + 5 = 0 \cr
  & \Rightarrow 2x + y + 5 = 0 \cr
  & \Rightarrow 2(X + 1) + Y + 2 + 5 = 0 \cr
  & \Rightarrow 2X + 2 + Y + 2 + 5 = 0 \cr
  &\Rightarrow 2X + Y + 9 = 0 \cr} $$
Compare this equation with (1)
Then,
$$\eqalign{
  & \Rightarrow 9(2k + 3) = 9k + 45 \cr
  &\Rightarrow 18k + 27 = 9k + 45 \cr
  & \Rightarrow 9k = 45 - 27 = 18 \cr
  & \Rightarrow k = 2 \cr} $$
Hence, value of $$k = 2$$

Note: We had a formula on basic of this question which is: if two lines are conjugate line with respect to circle $x^2+y^2=r^2$ then that satisfies certain equation which is $${r^2}({l_1}{l_2} + {m_1}{m_2}) = {n_1}{n_2}$$ . we have given two lines and also a circle equation. After equating both the equations, we got the value of k, which is 2.