If the portion of the linelx+my=1 falling inside the circlex2+y2=a2 subtends an angle of 45° at the origin, then
4[(a2(l2+m2)–1)]=a2(l2+m2)
4[(a2(l2+m2)–1)]=a2(l2+m2)-2
4[(a2(l2+m2)–1)]=[a2(l2+m2)-2]2
None of these
Step1: Simplify the given equations,
Given the equation of a line is lx+my=1
⇒ (lx+my)1=1..(i)
Equation of circle is x2+y2=a2…(ii)
Homogenizing (i)
x2+y2–a2(1)2=0
⇒ x2+y2–a2(lx+my)2=0
⇒ x2(1–a22)–2lma2xy+y2(1–a2m2)=0…(iii)
This is of the form Ax2+2hxy+By2=0
tanθ=2√(H2–AB)(A+B)
⇒ tan45=2√(H2–AB)(A+B)
⇒ 1=2√(H2–AB)(A+B)
⇒ (A+B)=2√(H2–AB)
From(iii)A=(1–a2l2),H=-lma2,B=(1–a2m2)
So 1–a2l2+1–a2m2=2√(l2m2a4–(1–a2l2)(1–a2m2))
2–a2(l2+m2)=2√(a2(l2+m2)–1)
Squaring both sides
(2–a2(l2+m2))2=4[(a2(l2+m2)–1)]
⇒4[(a2(l2+m2)–1)]=(a2(l2+m2)-2)2
Hence option C is the answer.
If the portion of the linelx+my=1 falling inside the circle x2+y2=a2 subtends an angle of 450 at the origin, then