The co-ordinates of the circumcentre of the triangle having vertices (-2, -3), (-1,0) and (7,-6), are (a) (3,-3) (b) (-3,-3) (c) (-3,3) (d) (3,3)

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(a) Let ( mathrm{S}(x, y) ) be the circumcentre of ( Delta P Q R ). ( Rightarrow quad S P=S Q=S R ) ( Rightarrow quad S P^{2}=S Q^{2}=S R^{2} ) Now, ( S P^{2}=S Q^{2} ) ( Rightarrow quad(x+2)^{2}+(y+3)^{2}=(x+1)^{2}+y^{2} ) ( Rightarrow quad 2 x+6 y+12=0 Rightarrow x+3 y+6=0 ldots ldots ) Consider ( S Q^{2}=S R^{2} ) ( Rightarrow quad(x+1)^{2}+y^{2}=(x-7)^{2}+(y+6)^{2} ) ( Rightarrow 16 x-12 y-84=0 ) ( Rightarrow quad 4 x-3 y-21=0 ) On solving (i) and (ii), we get ( x=3, y=-3 ) ( therefore ) Circumcentre ( =(3,-3) )

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