Question

If the portion of the line$lx+my=1$ falling inside the circle$x^2+y^2=a^2$ subtends an angle of $45°$ at the origin, then

• A. $4\left[\right(a^2(l^2+m^2)–1\left)\right]=a^2(l^2+m^2)$
• B. $4\left[\right(a^2(l^2+m^2)–1\left)\right]=a^2(l^2+m^2)-2$
• C. $4\left[\right(a^2(l^2+m^2)–1\left)\right]={\left[a^2\right(l^2+m^2)-2]}^2$
• D. None of these

Answer 1

The correct option is C

$4\left[\right(a^2(l^2+m^2)–1\left)\right]={\left[a^2\right(l^2+m^2)-2]}^2$

Step1: Simplify the given equations,

Given the equation of a line is $lx+my=1$

$\Rightarrow$ $\frac{(lx+my)}{1}=1..\left(i\right)$

Equation of circle is $x^2+y^2=a^2\dots \left(ii\right)$

Homogenizing (i)

$x^2+y^2–a^2{\left(1\right)}^2=0$

$\Rightarrow$ $x^2+y^2–a^2{(lx+my)}^2=0$

$\Rightarrow$ $x^2(1–\frac{a^2}{2})–2lma2xy+y^2(1–a^2{m}^2)=0\dots \left(iii\right)$

This is of the form $A{x}^2+2hxy+B{y}^2=0$

$\tan \theta =2\frac{\surd (H2–AB)}{(A+B)}$

$\Rightarrow$ $\tan 45=2\frac{\surd (H2–AB)}{(A+B)}$

$\Rightarrow$ $1=2\frac{\surd (H2–AB)}{(A+B)}$

$\Rightarrow$ $(A+B)=2\surd (H2–AB)$

From$\left(iii\right)A=(1–a^2{l}^2),H=-lm{a}^2,B=(1–a^2{m}^2)$

So $1–a^2{l}^2+1–a^2{m}^2=2\surd (l^2{m}^2{a}^4–(1–a^2{l}^2\left)\right(1–a^2{m}^2))$

$2–a^2(l^2+m^2)=2\surd \left(a^2\right(l^2+m^2)–1)$

Squaring both sides

${(2–a^2(l^2+m^2\left)\right)}^2=4\left[\right(a^2(l^2+m^2)–1\left)\right]$

$\Rightarrow$$4\left[\right(a^2(l^2+m^2)–1\left)\right]={(a^2(l^2+m^2)-2)}^2$

Hence option C is the answer.