Question

Find the circumcenter of a triangle whose sides are given by x+y+2=0,5x-y-2=0,x-2y+5=0

Answer 1

Answer:

$( - 9,18)$

Step-by-step explanation:

∆ Start by writing the Equation for circumcircle, so that you can pretty easily determine the center.

∆ Equation for circumcircle is given by :

$C : \alpha L_1L_2 + \beta L_2L_3 + \gamma L_3L_1 = 0$

Note that all three vertices of the ∆ satisfies the above equation.

✓ Your task is to equate the coefficients of squares of x and y and to remove any term of ( xy )

$\implies 5 \alpha + 5 \beta + \gamma = - \alpha + 2 \beta - 2 \gamma$

Additionally,

$4 \alpha - 11 \beta - \gamma = 0 \: (removing \: all \: terms \: of \: xy )$

$put \: \alpha = 5 \\ \\ \implies \: \beta = 3 , \gamma = - 13$

Hence, our Equation for Circumcircle is :

$C : 5 L_1L_2 + 3 L_2L_3 - 13L_3L_1 = 0$

And therefore, the center will be :

$( - \frac{ 8 \alpha + 23 \beta + 7 \gamma }{2} , \frac{4 \alpha + \beta - \gamma }{2} )$

_________________________________________

✓ That's one way of doing it.

✓ You can also try finding the vertices, writing a family through two vertex and making it pass through the third and hence find the circumcircle...

∆ But at the end,

^_^ It's all about Simplicity

Answer 2

Step-by-step explanation:

∆ Start by writing the Equation for circumcircle, so that you can pretty easily determine the center.

∆ Equation for circumcircle is given by :

C : \alpha L_1L_2 + \beta L_2L_3 + \gamma L_3L_1 = 0C:αL

1

L

2

+βL

2

L

3

+γL

3

L

1

=0

Note that all three vertices of the ∆ satisfies the above equation.

✓ Your task is to equate the coefficients of squares of x and y and to remove any term of ( xy )

\implies 5 \alpha + 5 \beta + \gamma = - \alpha + 2 \beta - 2 \gamma⟹5α+5β+γ=−α+2β−2γ

Additionally,

4 \alpha - 11 \beta - \gamma = 0 \: (removing \: all \: terms \: of \: xy )4α−11β−γ=0(removingalltermsofxy)

\begin{gathered}put \: \alpha = 5 \\ \\ \implies \: \beta = 3 , \gamma = - 13\end{gathered}

putα=5

⟹β=3,γ=−13

Hence, our Equation for Circumcircle is :

C : 5 L_1L_2 + 3 L_2L_3 - 13L_3L_1 = 0C:5L

1

L

2

+3L

2

L

3

−13L

3

L

1

=0

And therefore, the center will be :

( - \frac{ 8 \alpha + 23 \beta + 7 \gamma }{2} , \frac{4 \alpha + \beta - \gamma }{2} )(−

2

8α+23β+7γ

,

2

4α+β−γ

)l