Question

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

• A. (10, 10)
• B. (10, 12)
• C. (12, 12)
• D. (15, 15)
• E. (17, 17)

Answer 1

Answer: (E)

Let the circumcentre of triangle be $P(x,y)$
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).