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 IInd 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).