Question

The circumcenter 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

Hint: The circumcenter is the center point of the circumcircle drawn around a polygon. The circumcircle of a polygon is the circle that passes through all of its vertices and the center of that circle is called the circumcenter. All polygons that have a circumcircle are known as cyclic polygons. Only regular polygons, triangles, rectangles, and right-kites can have the circumcircle and thus the circumcenter.

Complete step-by-step answer:
Steps to construct the circumcenter of a triangle:
Step 1: Draw the perpendicular bisectors of all the sides of the triangle using a compass.
Step 2: Extend all the perpendicular bisectors to meet at a point. Mark the intersection point as O, this is the circumcenter.
Step 3: Using a compass and keeping O as the center and any vertex of the triangle as a point
on the circumference, draw a circle, this circle is our circumcircle whose center is O.

Assume that the circumcenter of a triangle is P(x,y)
The vertices are given to us as follows
A $$(0,30)$$
B $$(4,0)$$
C $$(30,0)$$
we have the following equations using the distance formula :
$$AP=\sqrt{{(x-0)}^2+{\left(y-30\right)}^2}$$
$$BP=\sqrt{{(x-4)}^2+{\left(y-0\right)}^2}$$
And
$$CP=\sqrt{{(x-30)}^2+{\left(y-0\right)}^2}$$
Since the vertices of the triangle are equidistant from the circumcenter .
Therefore we get
AP=BP=CP
Now using the first and second equality we have
$$\sqrt{{(x-0)}^2+{\left(y-30\right)}^2}$$ = $$\sqrt{{(x-4)}^2+{\left(y-0\right)}^2}$$
Squaring both the sides we get
$$(x-0{)}^2+{\left(y-30\right)}^2$$ = $$(x-4{)}^2+{\left(y-0\right)}^2$$
Which simplifies to
$$x^2+y^2+900-60y=x^2+16-8x+y^2$$
On further simplification we get
$$900+60y=16+8x$$
On further simplification we get
$$221=15y-2x$$ …(1)
 Now using the second and third equality we get
$$\sqrt{{(x-4)}^2+{\left(y-0\right)}^2}=\sqrt{{(x-30)}^2+{\left(y-0\right)}^2}$$
Squaring both the sides we get
$$(x-4{)}^2+{\left(y-0\right)}^2=(x-30{)}^2+{\left(y-0\right)}^2$$
Which simplifies to
$$x^2+16-8x+y^2=x^2+900-60x+y^2$$
On further simplification we get
$$884=52x$$
Therefore we get
$$x=17$$
Now putting this value of $$x$$ in (1) we get
$$y=17$$
Therefore the point of circumcentre $$(17,17)$$
Therefore option (E) is the correct answer.
So, the correct answer is “Option E”.

Note: Properties of circumcenter are: Consider any triangle ABC with circumcenter O.
A) All the vertices of the triangle are equidistant from the circumcenter.
B) All the new triangles formed by joining O to the vertices are Isosceles triangles.