Solveeit Logo

Question

Question: Find value of \[k\] if \[kx + 3y - 1 = 0,2x + y + 5 = 0\] are conjugate lines with respective to cir...

Find value of kk if kx+3y1=0,2x+y+5=0kx + 3y - 1 = 0,2x + y + 5 = 0 are conjugate lines with respective to circle x2+y22x4y4=0{x^2} + {y^2} - 2x - 4y - 4 = 0

Explanation

Solution

If two lines are conjugate line with respect to circle x2+y2=r2x^2+y^2=r^2 then that satisfies certain equation which is r2(l1l2+m1m2)=n1n2{r^2}({l_1}{l_2} + {m_1}{m_2}) = {n_1}{n_2}. Here we have equate given two lines with l1x+m1y+n1=0{l_1}x + {m_1}y + {n_1} = 0 and l2x+m2y+n2=0{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 r2(l1l2+m1m2)=n1n2{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 l1x+m1y+n1=0{l_1}x + {m_1}y + {n_1} = 0 and l2x+m2y+n2=0{l_2}x + {m_2}y + {n_2} = 0 are conjugate line with respect to circle x2+y2=r2{x^2} + {y^2} = {r^2} . In this condition it satisfies the equation r2(l1l2+m1m2)=n1n2{r^2}({l_1}{l_2} + {m_1}{m_2}) = {n_1}{n_2} …….. (1)
We have equations kx+3y1=0,2x+y+5=0kx + 3y - 1 = 0,2x + y + 5 = 0 and x2+y22x4y4=0{x^2} + {y^2} - 2x - 4y - 4 = 0
We can reform the circle equation as (x1)2+(y2)2=9{(x - 1)^2} + {(y - 2)^2} = 9 now we have to convert the equation (x1)2+(y2)2=9{(x - 1)^2} + {(y - 2)^2} = 9 in x2+y2=r2{x^2} + {y^2} = {r^2} form, for that we have to consider X=x1,Y=y2X = x - 1,Y = y - 2 then circle equation will be X2+Y2=9{X^2} + {Y^2} = 9
Put values of xx and yy in given equation kx+3y1=0,2x+y+5=0kx + 3y - 1 = 0,2x + y + 5 = 0
Equation will become

& 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.