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  • Find the centre of the circum-circle of the triangle formed by the pair of lines <img alt="\mathrm{7 x^2+8 x y-y^2=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B7%20x%5E

Find the centre of the circum-circle of the triangle formed by the pair of lines \mathrm{7 x^2+8 x y-y^2=0} and the line \mathrm{2x+y=1}.

Option: 1

(\frac{-12}{13},\frac{-4}{13})


Option: 2

(\frac{12}{13},\frac{4}{13})


Option: 3

(\frac{25}{13},\frac{4}{13})


Option: 4

None of these 


Answers (1)

best_answer

The pair of lines given by \mathrm{7 x^2+8 x y-y^2=0} pass through the origin. One of the vertices of the triangle, therefore, is the origin. Hence the circum-cirle passes through the origin. Let the equation of the circle be \mathrm{x^2+y^2+2 g x+2 f y=0}. For the equation of the pair of lines passing through the intersection of this circle with the line \mathrm{2x + y = 1} and through the origin, we homogenise the equation of the circle with the help of the line \mathrm{2x + y = 1}.

\mathrm{\Rightarrow x^{2} + y^{2} + 2gx(2x + y) + 2fy(2x + y) = 0}

which is identical to \mathrm{7x^{2} + 8xy - y^{2} = 0}

\mathrm{\Rightarrow \frac{7}{1+4 g}=\frac{-1}{1+2 f}=\frac{8}{2 g+4 f} \Rightarrow g=-\frac{12}{13}, f=-\frac{4}{13} \text {. }}

Hence the equation of the required circle is \mathrm{13 x^2+13 y^2-24 x-8 y=0}

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Nehul

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