question_answer

The radical centre of three circles described on the three sides of a triangle as diameter is                          [EAMCET 1994]

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.