Find the circumcenter of the triangle
whose sides are given by x + y + 2 = 0,
5x - y - 2 = 0 and x - 2y + 5 = 0
Answers
Answer:
∆ 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
)
_________________________________________
✓ 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,