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  • If cyclic quadrilateral is formed by the lines  <img alt="\mathrm{5 x+3 y=9, x=3 y, 2 x=y}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B5%20x&plus;3%20y%3D9%2C%20

If cyclic quadrilateral is formed by the lines  \mathrm{5 x+3 y=9, x=3 y, 2 x=y}  and  \mathrm{x+4 y+2=0}  taken in order, then the equation of the circum-circle is

Option: 1

\mathrm{9\left(x^2+y^2\right)-10 x+15 y=0 }


Option: 2

\mathrm{9\left(x^2+y^2\right)-20 x+15 y=0 }


Option: 3

\mathrm{3\left(x^2+y^2\right)-20 x+15 y=0 }


Option: 4

\mathrm{3\left(x^2+y^2\right)-10 x+5 y=0}


Answers (1)

best_answer

\mathrm{A\left(\frac{3}{2}, \frac{1}{2}\right), B(0,0), C\left(-\frac{2}{9},-\frac{4}{9}\right), D\left(\frac{42}{17},-\frac{19}{17}\right) \text {. }}

The equation of the circle is \mathrm{x^2+y^2+2 g x+2 f y=0}  as passes through B(0,0)
It passes through A and C
\mathrm{\therefore \frac{9}{4}+\frac{1}{4}+3 g+f=0 \Rightarrow 10+12 g+4 f=0}
and   \mathrm{\frac{4}{81}+\frac{16}{81}-\frac{4 g}{9}-\frac{8 f}{9}=0 \Rightarrow 20-36 g-72 f=0}
Solving for f and g, we get   \mathrm{g=-\frac{10}{9}, f=\frac{5}{6}}
Hence the circle is    \mathrm{9\left(x^2+y^2\right)-20 x+15 y=0}.

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vishal kumar

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