The equation of the circle circumscribing the triangle formed by the straight lines \(x+y=6\). \(2 x+y=4\)…
The equation of the circle circumscribing the triangle formed by the straight lines \(x+y=6\). \(2 x+y=4\) and \(x+2 y=5\) is given by
\(x^2+y^2+17 x+19 y+50=0\)
\(x^2+y^2-17 x-19 y+50=0\)
\(x^2+y^2+17 x-19 y-50=0\)
\(x^2+y^2-17 x+19 y-50=0\)
Solution
Equation of circle circumscribing the triangle formed by the straight lines, \(x+y=6,2 x+y=4\) and \(x+2 y=5\) is
\(\begin{aligned}
& (x+y-6)(2 x+y-4)+\lambda(2 x+y-4)(x+2 y-5) \\
& +\mu(x+2 y-5)(x+y-6)=0
\end{aligned}\)
\(\because\) For a circle the coefficients of terms \(x^2\) and \(y^2\) must be equal and non-zero and coefficient of \(x y\) must be zero.
\(\therefore\) Coefficient of \(x^2=\) coefficient of \(y^2\)
\(\Rightarrow 2+2 \lambda+\mu=1+2 \lambda+2 \mu \Rightarrow \mu=1\)
and coefficient of \(x y=0\)
\(\Rightarrow 1+2+4 \lambda+\lambda+\mu+2 \mu=0 \Rightarrow \lambda=-6 / 5 \text {. }\)
So, equation of required circle is
\(\begin{array}{ll}
(x+y-6)(2 x+y-4)-\frac{6}{5}(2 x+y-4)(x+2 y-5)+(x+2 y-5)(x+y-6)=0 \\
\Rightarrow 5(x+y-6)(3 x+3 y-9)-6(2 x+y-4)(x+2 y-5)=0 \\
\Rightarrow 5(x+y-6)(x+y-3)-2(2 x+y-4)(x+2 y-5)=0
\end{array}\)
\(\begin{aligned}
\Rightarrow & 5\left(x^2+x y-3 x+x y+y^2-3 y-6 x-6 y+18\right) \\
& -2\left(2 x^2+4 x y-10 x+x y+2 y^2-5 y-4 x\right. \\
\Rightarrow & x^2+y^2-45 x-45 y+90+28 x+26 y-40=0 \\
\Rightarrow & x^2+y^2-17 x-19 y+50=0
\end{aligned}\)
Hence, option (b) is correct.