106f(x,y) = x2 + y2 + 2gx+2fy +c
f(x,1) = x2 + 1 + 2gx+2f+c ≡ x2 - 4x +4
=> -4 = 2g .. (i)
4 = 1+2f+c ... (ii)
f(1,x) = 1+x2 + 2g+2fx+c ≡ x2
=> 2fx =0 => f=0
and
1+2g+c = 0 .. (iii)
solve these three and ur done
Ques) Suppose f(x,y) = 0 is the equation of the circle such that f(x,1) = 0 has equal roots (Each equals to 2) and f(1,x) = 0 also has equal roots (Each equals to 0 ). Then find the equation of the circle.
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UP 0 DOWN 0 0 2
2 Answers
106
11Oh Yeah, thanx a lot Asish
This ques looks difficult, but u made this easy. Thanx a lot