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  • Find the equation of the circum-circle of the triangle formed by the pair of lines <img alt="7 x^2+8 x y-y^2=0" src="https://entrancecorner.oncodecogs.com/gif.latex?7%20x%5E2&plus;8%20x%20y-y%5

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

Option: 1

x^2+y^2-24 x-8 y=0


Option: 2

13x^2+13y^2-24 x-8 y=0


Option: 3

13x^2+13y^2-8 x-4 y=0


Option: 4

x^2+y^2-2 x- y=0


Answers (1)

best_answer

The pair of lines given by 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 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 2 x+y=1 and through the origin, we homogenise the equation of the circle with the help of the line 2 x+y=1.
\Rightarrow x^2+y^2+2 g x(2 x+y)+2 f y(2 x+y)=0
which is identical to 7 x^2+8 x y-y^2=0
\begin{aligned} & \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} . \end{aligned}
Hence the equation of the required circle is 13 x^2+13 y^2-24 x-8 y=0

Posted by

Gautam harsolia

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