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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 Mar 2007 22:23:05 IST
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the equation of a circle which touches the axis of y at a distance 4 from the origion & cuts off an intercept 6 from the axis of x???
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9 Mar 2007 22:24:46 IST
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pls reply me immediately as it is very urgent i m too confused
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9 Mar 2007 23:11:19 IST
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Hi Deep, Let ( g , f ) be center of circle. Y axis is tangent to circle, therefore f = + - 4 X axis is intercepted by circle . So , 2(g^2 - c )^(1/2) = 6 ____________(i) and intercept on Y axis is 0 . So, 2(f^2 - c)^ (1/2) = 0 f^2 = c 16 = c From (i) g = + - 5 radius is 5 radius = ( g^2 + f^2 - c) ^ (1/2) ( x + - 5 )2 + (y + - 4 )2 = 25 
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Think different .........................think apple
ALBERT EINSTEIN |
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9 Mar 2007 23:16:39 IST
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sorry the answer is wrong
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9 Mar 2007 23:34:24 IST
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hey deep whats wrong with my answer u can crosscheck
If u substitute (0 , - + 4) in equation it will satisfy.
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Think different .........................think apple
ALBERT EINSTEIN |
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10 Mar 2007 00:06:52 IST
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even i got the same solution... n i cant understand why you put this as 'a good solution' and why this answer is wrong!!
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Aashii, Simply can't study for long |
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10 Mar 2007 00:40:53 IST
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HELLO DEEP 01................... VINEET HAS DONE IN A VERY GOOD MANNER............ BUT THERE IS ONE MISTAKE.......... I WILL SOLVE IT......... | Let ( g , f ) be center of circle. f = 4 X axis is intercepted by circle . So , 2(g^2 - c )^(1/2) = 6 ____________(i) and intercept on Y axis is 0 . So, 2(f^2 - c)^ (1/2) = 0 f^2 = c 16 = c From (i) g = + - 5 THEREFORE radius = sqrt ( g2 + f2 - c ) = 3 now look....................... one more thing................ 4 circles will be drawn...............okkkkkkkkkkkk............... and............their centre will be.......... (5,4) (-5,4) (5,-4) (-5,-4) radius same =3......... i m writing 4 only circle in same quadrant........... (x-5)2 + ( y-4)2 = 9.............. it will be right ................. thanku......... | |
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BELIEVE IN URSELF.....BECAUSE I BELIEVE IN U.............. |
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10 Mar 2007 02:12:42 IST
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well apply family of circles...
the equation is in terms of k-
(x)^2 + (y-4)^2 + k(x)=0
and x intercept =6
hence 2(K/2^2-16)=6
hence k=+- 10
and hence eq is-
x^2+y^2+10x-8y+16=0
or
x^2+y^2-10x-8y+16=0
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13 Mar 2007 19:05:59 IST
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hey ankur, i think your initial equation shud hav (y+-4)^2 since da circle can also be in da third or fourth quadrant wich will add two more solutions 2 ur answer
x^2 + y^2 -10x + 8y + 16 = 0
&
x^2 + y^2 +10x + 8y +16 = 0
wich is same as da answer as given by vineet
n DON007, how did u get the square root of g^2 +f^2 - c = 3... it comes out to be 5 wich is again da same answer
deep01 please step up n say y is dis answer wrong
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Aashii, Simply can't study for long |
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13 Mar 2007 19:31:06 IST
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hey guys(and a gal) ,dont u think if the circle touches the y-axis at a distance 4 ,its radius = 4 i got the centre as (+ - 4, + - sqroot(7) ) there4 equation of circle is x^2 + y^2 +-8x +-2sqroot(7)y + 7 = 0 is this correct ?
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light travels the fastest ??? NO
wherever light goes it always finds that darkness has already got there |
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13 Mar 2007 19:48:03 IST
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let the eqn be x^2 + y^2 + 2gx + 2fy + c = 0
The circle passes thru (0.4) ---
16 + 16f +c = 0 ---i
Circle has 0 intercept on y axis -----
f^2 = c ---------------------- ii
Circle has 6 intercept on x axis -----
g^2 - c = 9
we get --- f = -15 or -1
c = 225
g = 15.29
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13 Mar 2007 19:56:48 IST
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hey deep
since the circle touches yaxis at (0,4)
=> the centre is (x,4)
also perpendicular from centre to the pppposite chord(the intercept of 6 on x axis)=4
just see a rt triangle is formed with the hypotenuse being the radius
thus r^2=3^2+ 4^2
=>r=5
also distance from centre to y axis =r=5
thus centre (5,4)
eq. (x-5)^2+ (y-4)^2=25
is it ok!
i think it is!!
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there is no right way 2 do something wrong !!!!!!!! |
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25 Mar 2007 13:09:21 IST
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