Suppose A and B are the points at which the line x+y−λ=0 meets the pair of straight lines x2+y2−2x−4y+2=0. If ∠AOB=90∘, then a value of λ is
=1 ...(i) and pair of straight lines is x2+y2−2x−4y+2=0 We will make homogeneous of the above line. x2+y2−2x(l)−4y(l)+2(l)2=0 ⇒x2+y2−2x(
x+y
λ
)−4y(
x+y
λ
) +2(
x+y
λ
)2=0 [using Eq. (i)] ⇒λ2(x2+y2)−2x2λ−2yxλ−4xyλ−4y2λ+2(x2+y2+2xy)=0 ⇒(λ2−2λ+2)x2+(4xy−6xyλ)(λ2−4λ+2)y2=0 ⇒(λ2−2λ+2)x2+(4−6λ)xy+(λ2−4λ+2)y2=0 Now, ∠AOB=90∘ where O is the origin. Condition for ∠ADB=90∘ is given by- (Coefficient‌of‌x2)+(Coefficient‌of‌y2)=0 ⇒(λ2−2λ+2)+(λ2−4λ+2)=0 ⇒2λ2−6λ+4=0 ⇒λ2−3λ+2=0 ⇒(λ−1)(λ−2)=0 λ=1,2 Hence, option (a) is correct.