A) The orthocentre
B) The circumcentre
C) The incentre of the triangle
D) The centroid
Correct Answer: A
Solution :
Let us consider a triangle as shown in fig. Equations of the circles with AB, BC and CA as diameters are \[{{S}_{1}}\equiv (x+a)(x-a)+{{y}^{2}}=0\] \[{{S}_{2}}\equiv (x-a)(x-\alpha )+y(y-\beta )=0\] and \[{{S}_{3}}\equiv (x+a)(x-\alpha )+y(y-\beta )=0\] i.e., \[{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}-{{a}^{2}}=0\] \[{{S}_{2}}\equiv {{x}^{2}}+{{y}^{2}}-(a+\alpha )x-\beta y+a\alpha =0\] and\[{{S}_{3}}\equiv {{x}^{2}}+{{y}^{2}}-(\alpha -a)x-\beta y-a\alpha =0\] \ Radical axis of \[{{S}_{2}}\] and \[{{S}_{3}}\] is \[{{S}_{3}}-{{S}_{2}}=0\] i.e., \[2ax-2a\alpha =0\] Þ \[2a(x-\alpha )=0\], as \[a\ne 0\], \[x=\alpha \] But \[x=\alpha \] is the orthogonal through C. Similarly other radical axes will be orthogonals through A and B. Hence radical centre will be the orthocentre.You need to login to perform this action.
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