A) (10, 10)
B) (10, 12)
C) (12, 12)
D) (15, 15)
E) (17, 17)
Correct Answer: E
Solution :
Let the circumcentre of triangle be\[P(x,\text{ }y)\]and let the vertices of a triangle be A (0, 30), B (4, 0) and C (30, 0). \[\therefore \] \[P{{A}^{2}}=P{{B}^{2}}=P{{C}^{2}}\] \[\Rightarrow \]\[{{(x-0)}^{2}}+{{(y-30)}^{2}}\] \[={{(x-4)}^{2}}+{{(y-0)}^{2}}\] \[={{(x-30)}^{2}}+{{(y-0)}^{2}}\] From Und and IIIrd terms, \[{{x}^{2}}+{{y}^{2}}-60y+900={{x}^{2}}+{{y}^{2}}-8x+16\] \[\Rightarrow \] \[8x-60y+884=0\] ...(i) From IInd and IIIrd terms, \[{{x}^{2}}-8x+16+{{y}^{2}}={{x}^{2}}-60x+900+{{y}^{2}}\] \[\Rightarrow \]\[52x=884\Rightarrow x=17\] On putting x = 17 in Eq. (i), we get \[y=17\] Hence, required point is (17, 17).
You need to login to perform this action.
You will be redirected in
3 sec
