Question

Find the value of k, if kx + 3y - 1 = 0, 2x + y + 5 = 0 are conjugate lines with respect to circle x^2 + y^2 - 2x - 4y - 4 = 0.

Ans : 2

Answer 1

Step-by-step explanation :-

Given that :

• Equation of circle is $\bold{x^2+y^2-2x-4y-4=0}$.

• Equation of line are $L_1 = \bold{kx+3y-1=0}\;,\,L_2 = \bold{2x+y+5=0}$.

Here :

• $\bold{l_1=k,m_1=3,n=-1,l_2=2,m_2=1,n_2=5}$

• $\bold{g=-1,f=-2,c=-4}$

So :

• $\bold{r=\sqrt{g^2+f^2-c}=\sqrt{1+4+4}=3 }$

∵ If the lines $\bold{l_1x+m_1y+n_1=0\;and\;l_2x+m_2y+n_2=0}$ are conjugate w..r. to the circle $\bold{S=x^2+y^2+2gx+2fy+c=0}$ then $\bold{r^2(l_1l_2+m_1m_2)=(l_1g+m_2f-n_1)(l_2g+m_2f-n_1)}$.

$\rightarrowtail \bold{9[k(2)+3(1)=[k(-1)+3(-2)+1][2(-1)+1(-2)-5]}$

$\rightarrowtail \bold{9(2k+3)=(-k-6+1)(-2-2-5)}$

$\rightarrowtail \bold{9(2k+3)=(-k-5)(-9)}$

$\rightarrowtail \bold{2k+3=k+5}$

$\rightarrow \bold{k=2}$

Answer 2

Concept:

The set of all points in the plane that are a fixed distance (the radius) from a fixed point is called a circle (the centre). A radius is the distance between any two points on a circle and the centre. The polar of P and Q are called conjugate lines with respect to the circle S = 0 if P and Q are conjugate points with respect to the circle S = 0.

Given:

Equation of circle:

$x^2+y^2-2x-4y-4=0\\g=-1, f=-2, c=-4$

Conjugate lines

$L_1:kx+3y-1=0\\l_1=k, m_1=3, n_1=-1\\L_2:2x+y+5=0\\l_2=2, m_2=1, n_2=5$

Find:

The value of k in conjugate lines.

Solution:

The radius of the circle:

$r=\sqrt{g^2+f^2-c} \\r=\sqrt{1+4+4}\\ r=3$

Condition for conjugate lines:

$r^2(l_1l_2+m_1m_2)=(l_1g+m_1f-n_1)(l_2g+m_2f-n_2)\\9(2k+3)=(-k-6+1)(-2-2-5)\\2k+3=k+5\\k=2$

Hence, the value of $k$ is $2$.

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