The line x+y=k meets the pair of straight lines x^2+y^2-2x-4y+2=0 at two points A and B. If O is the origin and ∠ AOB=90^ , then the value of k (1 ) is

Question

The line x+y=k meets the pair of straight lines x^2+y^2-2x-4y+2=0 at two points A and B. If O is the origin and ∠ AOB=90^ , then the value of k (>1 ) is

Accepted Answer

Correct answer: D. (d) Given two-degree equation, x^2+y^2-2x-4y+2=0 and given line, L ≡x+y=k ⇒L= ( x+yk ) To make homogenious, the pair of lines x^2+y^2-2x (1 )-4y (1 )+2 ( x+yk )^2=0 We get x^2+y^2-2x ( x+yk )-4y ( x+yk )+2 ( x+yk )^2=0 Since, intersection points of line and pair of lines make an angle 90^ at origin O. ∴ Coefficient of x^2+ Coefficient of y^2=0 ⇒(1- 2k+ 2k^2)+(1- 4k+ 2k^2)=0 ⇒k^2-3k+2=0 ⇒ (k-2 ) (k-1 )=0 ∵k>1 so ,k=2

$The line x + y = k meets the pair of straight lines x^{2} + y^{2} - 2 x - 4 y + 2 = 0 at two points A and B . If O is the origin and ∠ AOB = 90^{\circ}, then the value of k (> 1) is$

Solution

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