11My answer is (-1,0)
method: i jus fnd the eqn of chord of contact and the foot of perpendicular from the centre would give the midpoint of chord of contact
find teh coordinates of point from which the tangents are drawn to teh circle x2+y2-6x-4y+3=0 such that the mid pt of chord of contact is (1,1)
-
UP 0 DOWN 0 0 5
5 Answers
11
62Eureka.. try this one by geometry again!!!!!!!!!!! (You will be amazed yourself.. that how often question can be solved on coordinate geometry.. just by using pure geometry!!)
Now Draw the cricle with center (3,2) and radius of √10
And now draw the chord with center (1,1)
Now draw the tangent to the ends of the chord... (there will be some angles that are 90 degrees!!!)
I hope this helps..
If this doesnt then i will post the final solution :)
1w.k.t the eqn of chord with m.p as (h,k) is given by T=S1 .
then , we've got to find that point , which has this as the chord of caontct . T=0 , I think we'll be better of substituing values as time is a factor. ( options are given no ? ) that's why . I felt this is easier but not now , in xam. :P