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Find the condition on \mathrm{ a, b, c} such that two chords of the circle
\mathrm{ x^2+y^2-2 a x-2 b y+a^2+b^2-c^2=0 }
passing through the point \mathrm{(a, b+c)} are bisected by the line \mathrm{y=x.}
 

Option: 1

\mathrm{4 a^2+4 b^2-4 c^2-8 a b+b c-a c<0}
 


Option: 2

\mathrm{4 a^2+4 b^2-8 a b+4 b c-4 a c-c^2<0}
 


Option: 3

\mathrm{4 a^2+2 b^2-4 a b+4 b c-a c+c^2>0}
 


Option: 4

\mathrm{4 a^2+4 b^2-8 a b-c^2>0}


Answers (1)

best_answer

Chords are bisected on the line \mathrm{y=x}. Let \mathrm{\left(x_1, x_1\right)} be the mid point of the chord then equation of the chord is \mathrm{\mathrm{T}=\mathrm{S}_1}

\mathrm{\therefore \quad \mathrm{xx}_1+\mathrm{yx}_1-\mathrm{a}\left(\mathrm{x}+\mathrm{x}_1\right)-\mathrm{b}\left(\mathrm{y}+\mathrm{x}_1\right)+\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2 }

\mathrm{ =\mathrm{x}_1^2+\mathrm{x}_1^2-2 \mathrm{ax}_1-2 \mathrm{bx}_1+\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2 }

\Rightarrow \quad \left(\mathrm{x}_1-\mathrm{a}\right) \mathrm{x}+\left(\mathrm{x}_1-\mathrm{b}\right) \mathrm{y}+\mathrm{ax}_1+\mathrm{bx}_1-2 \mathrm{x}_1^2=0

This chord passes through \mathrm{(a, b+c)}

\Rightarrow \quad\left(\mathrm{x}_1-\mathrm{a}\right) \mathrm{a}+\left(\mathrm{x}_1-\mathrm{b}\right)(\mathrm{b}+\mathrm{c})+\mathrm{ax}_1+\mathrm{bx}_1-2 \mathrm{x}_1^2=0

\Rightarrow \quad 2 \mathrm{x}_1^2-(2 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}) \mathrm{x}_1+\mathrm{a}^2+\mathrm{b}^2+\mathrm{bc}=0

which is quadratic in \mathrm{x_1}. Since it is given that two chords are bisected on the line \mathrm{\mathrm{y}=\mathrm{x}, then\: \: \mathrm{x}_1} must have two distinct real roots,

\mathrm{\therefore B^2-4 A C>0 }

\mathrm{\Rightarrow (2 a+2 b+c)^2-4 \cdot 2 \cdot\left(a^2+b^2+b c\right)>0 }

\mathrm{\Rightarrow 4 a^2+4 b^2+c^2+8 a b+4 b c+4 a c-8 a^2-8 b^2-8 b c>0 }

\mathrm{\Rightarrow 4 a^2+4 b^2-8 a b+4 b c-4 a c-c^2<0}

Hence the condition on \mathrm{a, b, c} is

\mathrm{ 4 a^2+4 b^2-c^2-8 a b+4 b c-4 a c<0 }

Hence option 2 is correct.
 

Posted by

manish

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