1Either i'm not getting what philip is saying or...I think he's right.
Find the range of values of λ for which the variable line 3x+4y - λ=0 lies between the circles:
x2+y2 - 2x - 2y +1=0
x2+y2 - 18x - 2y +78 = 0
without intercepting a chord on either circle.
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18 Answers
62(3+4-λ)(27+4-λ)<0
(7-λ)(31-λ)<0
so 7<λ<31
equality will hold or not depending on whether or not the touching of the circles is allowed..
|(3+4-λ)/5| >1
|7-λ| >5
7-λ > 5 so 2>λ
or 7-λ < -5 so 12<λ
|(31-λ)|/5 > 2
|(31-λ)| > 10
(31-λ) > 10 or λ<21
or (31-λ) < -10 or λ>41
The intersection of the threee ranges gives us [12, 21]
Which gives us the answer :)
1Ok so here is the solution
|7-λ| > 5 or λ-7>5
& |31-λ| > 10 or 31- λ > 10
=> 12 < λ < 21
seems so simple when we compare it to the conventional way
now the trivial thing to be observed is 7-λ is -ive and 31-λ is +ive.
once someone explains this the thread is closed.
remember that the explanation will be just a one liner....
11@ nishant bhaiya ...
did you read the hidden part of my post #12 ... that is this thing + much more...
1@ bhaiyya i have checked all sufficient conditions (the same 3 you pointed out in #2) and possibly you have checked some extra condition that you are getting so many different inequalities :P
if you still think otherwise please point out next time you see this
62no philip..
you have not checked the third condition..
it is only a coincidence that the third condition fell into place automatically
1@bhaiyya i still like my solution more......
there have been innumerable posts where you have had me mesmerised but this time yours was probably the more beaten method of doing this
62this one seems simple
first circle has a center of 1,1 and radius of 1
2nd one has a center of 9, 1 and a radius of 2
so you want the distance of the line from 1st circle to be >1
while from the 2nd to be > 2
moreover you want the signs to be opposite of each other so that the circles lie on either side of the line....
or otherwise you can check the position of the centers from the line .. they should be on either side...
1yes
ok let me write the 1st two...
|7-λ|5>1
and
|31-λ|5>2
now pls explain how to fit your step into this
1still...bhaiyya the answer should come out to be λ ε [12,21]
i dont see how your inequality will contribute then..
62wait i will post the complete solution and then you can see...! !
If ofcourse you dont mean to say that the step that i have given is wrong!
62kyun bhai..
I have obviously not checked the first two conditions.. The above result is subject ot checking the first condition that i stated!
62(3+4-λ)(27+4-λ)<0
(7-λ)(31-λ)<0
so 7<λ<31
equality will hold or not depending on whether or not the touching of the circles is allowed..
1yes
it is only the neatness with which someone can carry out those last two lines of you post that will matter...