1NOT LIKE DAT...
the given thing is....
x^2-y^2-2x+4y-3=0
=>( x^2 -2x + 1 ) - (y^2 - 4y +4) =0
=> (x-1)^2 - (y-2)^2=0
=> (x-1+y-2)(x-1-y+2)=0
............i hope its clear now...
the eqn of circle passing through (1,1) and having two diameters along the pair of lines x2-y2-2x+4y-3=0
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5 Answers
1the pair of srt lines eqn bcums : (x-1)^2-(y-2)^2=0
u will get two eqns: x+y-3=0 n x-y+1=0
solve these.
get center...
in eqn (x-h)^2+(y-k)^2 = r^2
put h,k value (center) n put (1,1) in place of x n y...
get value of r... :)
1357You have to solve the quadratic equation in terms of y (or x) to get two separate straight lines
1i think the st lines can be arrived at by partial differentiating the equation once with respect to x and again with respect to y
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