Let 2x2+y2−3xy=0 be the equation of a pair of tangents drawn from the origin O to a circle of radius 3 units with centre in the first quadrant. If A is one of the points of contact, then the length of OA is
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Solution
y2−3xy+2x2=0
Let θ be the angle between the two lines. Then, tanθ=2√(9/4)−21+2=13
but from angle between the formulae tanθ=2LrL2−r2,
Hence, 13=6LL2−9 ⇒L2−18L−9=0 ⇒L=9±√90
As L can't be negative, L=9+√90 ⇒L=3(3+√10) ∴OA=L=3(3+√10) units