The circumcentre of the triangle with vertices $(0,30),(4,0)$ and $(30,0)$ is

The correct option is E

$(17,17)$

Explanation for the correct option:

Finding the circumcentre:

Step 1: Equating circumradius to find the relation between $\left(x,y\right)$

Let $A(0,30),B(4,0)$ and $C(30,0)$ be vertices of a triangle.

And its circumcentre be $P(x,y)$.

Hence, ${\mathrm{PA}}^2={\mathrm{PB}}^2={\mathrm{PC}}^2$

Consider ${\mathrm{PA}}^2={\mathrm{PB}}^2$.

$\Rightarrow {(x-0)}^2+{(y-30)}^2={(x-4)}^2+{(y-0)}^2={(x-30)}^2+{(y-0)}^2$

Now,

$$\begin{gathered} x^2+y^2-60y+900=x^2+y^2+16-8x \\ \Rightarrow 8x-60y+884=0\dots \dots \left(1\right) \end{gathered}$$

Step 2: Equating circumradius to find the relation between $\left(x,y\right)$ using ${\left(PB\right)}^2={\left(PC\right)}^2$

Similarly,

$$\begin{gathered} x^2-8x+16+y^2=x^2-60x+900+y^2 \\ \Rightarrow 52x=883 \\ \Rightarrow x=17 \end{gathered}$$

Step 3: Using the substitution method to find the value $\left(x,y\right)$

Substituting the value of $x$ in equation $\left(1\right)$.

$$\begin{gathered} \therefore 8\left(17\right)-60y+884=0 \\ \Rightarrow 136+884=60y \\ \Rightarrow y=17 \end{gathered}$$

Hence, the coordinates of the circumcentre is $(17,17)$.

Therefore, option (E) is the correct answer.