The circumcircle of the triangle whose sides are
$L_{1} = 3 x + y - 5 = 0, L_{2} = 2 x + y - 3 = 0, L_{3} = 3 x + 2 y - 7 = 0$ is

x^{2} + y^{2} - 21 x + 7 y - 40 = 0

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x^{2} + y^{2} + 21 x - 7 y - 40 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} + 21 x + 7 y - 40 = 0

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x^{2} + y^{2} + 21 x + 7 y + 40 = 0

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Solution

The correct option is C $x^{2} + y^{2} + 21 x + 7 y - 40 = 0$
The desired circle is $α L_{2} L_{3} + β L_{3} L_{1} + γ L_{1} L_{2} = 0$
$\Rightarrow α (2 x + y - 3) (3 x + 2 y - 7) + β (3 x + 2 y - 7) (3 x + y - 5) + γ (3 x + y - 5) (2 x + y - 3) = 0$ ................... $(1)$
coeff. $x^{2} =$ coeff. $y^{2}$
$\Rightarrow 6 α + 9 β + 6 γ = 2 α + 2 β + γ$
$∴ 4 α + 7 β + 5 γ = 0$ (on transposing from RHS to LHS)
Coeff. $x y = 0 \Rightarrow 7 α + 9 β + 5 γ = 0$ .............. $(2)$
From eqns $(1)$ and $(2)$ we have
$\frac{α}{- 10} = \frac{β}{15} = \frac{γ}{- 13}$
or $α : β : γ = 10 : - 15 : 13$
Substituting for $α, β, γ$ in eqn $(1)$ we get
$10 (2 x + y - 3) (3 x + 2 y - 7) - 15 (3 x + 2 y - 7) (3 x + y - 5) + 13 (3 x + y - 5) (2 x + y - 3) = 0$
On simplification, we get
$x^{2} + y^{2} + 21 x + 7 y - 40 = 0$

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The circumcircle of the triangle whose sides areL1=3x+y−5=0,L2=2x+y−3=0,L3=3x+2y−7=0 is

The circumcircle of the triangle whose sides are
$L_{1} = 3 x + y - 5 = 0, L_{2} = 2 x + y - 3 = 0, L_{3} = 3 x + 2 y - 7 = 0$ is