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```        if 4l^2-5m^2+6l+1=0 then the line lx+my+1=0 touches a fixed circle of centre (3,0) and radius
a)3^1/2 b)2^1/2 c)5^1/2 d)7^1/2```
7 years ago

147 Points
```										Dear Vaibhav
4l2-5m2+6l+1=0
5( l2 +m2) =9l2 +6l +1
or 5( l2 +m2) =(3l+1)2
or v5v( l2 +m2) =|(3l+1)|
now length of perpendicular from(3,0)  on line lx +my +1 =0
r = |3l+1|/v( l2 +m2)
from above rresult
r = v5
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7 years ago
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