Question

The diameter of the nine point circle of the triangle with vertices $(3, 4)$, $\left( {5\cos \theta ,5\sin \theta } \right)$ and $\left( {5\cos \theta , - 5\sin \theta } \right)$ where $\theta \in {\mathbf{R}}$, is
A) 10
B) 2
C) 3
D) 5

Answer 1

Hint:
 By the property of nine point circle, we know that the radius ${r_n} = \dfrac{1}{2}R$ where, ${r_n}$ is the radius of the nine point circle and R is the radius of the circumcircle. Then, we will draw the triangle with the given vertices. Let the circumcentre, O be at the origin then, we will calculate OA, OB and OC to prove that they are equal. This implies that the distance of the vertices is equal from the circumcentre O and therefore, OA = OB = OC = radius of the circumcircle = R. then, we will put the value of R in ${r_n}$ and we know that the diameter of a circle is twice its radius.

Complete step by step solution:
We are given the vertices of a nine point circle as (3, 4), $\left( {5\cos \theta ,5\sin \theta } \right)$ and $\left( {5\cos \theta , - 5\sin \theta } \right)$ where $\theta \in {\mathbf{R}}$
We are required to find the diameter of the nine point circle.
We know that the radius of the nine point circle is half of the radius of the circumcircle i.e., ${r_n} = \dfrac{1}{2}R$ where, ${r_n}$ is the radius of the nine point circle and R is the radius of the circumcircle.
Now, let us draw the figure:

 $\left( {5\cos \theta ,5\sin \theta } \right)$ $\left( {5\cos \theta , - 5\sin \theta } \right)$

Here, the triangle ABC is represented in green and the red circle is the circumcircle of the triangle. The blue circle represents the nine point circle which passes through nine concyclic points defined from the triangle.
Now, let us calculate the distances OA, OB and OC using the distance formula: $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
$ \Rightarrow $ OA = $\sqrt {{{\left( {3 - 0} \right)}^2} + {{\left( {4 - 0} \right)}^2}} = \sqrt {9 + 16} = \sqrt {25} = 5$
Similarly,
OB = $\sqrt {{{\left( {5\cos \theta - 0} \right)}^2} + {{\left( {5\sin \theta - 0} \right)}^2}} = \sqrt {25{{\cos }^2}\theta + 25{{\sin }^2}\theta } = \sqrt {25\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)} = \sqrt {25} = 5$
And,
OC= $\sqrt {{{\left( {5\cos \theta - 0} \right)}^2} + {{\left( { - 5\sin \theta - 0} \right)}^2}} = \sqrt {25{{\cos }^2}\theta + 25{{\sin }^2}\theta } = \sqrt {25\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)} = \sqrt {25} = 5$
Here, we can see that OA = OB = OC. Therefore, the radius of the circumcircle is:
$ \Rightarrow $R = OA= OB = OC = 5
Now, by the property ${r_n} = \dfrac{1}{2}R$, we get
$ \Rightarrow {r_n} = \dfrac{1}{2}R = \dfrac{5}{2}$
We know that the diameter of a circle is twice its radius, therefore, the diameter of the nine point circle = 2${r_n}$ = 2$ \times \dfrac{5}{2}$ = 5.

Therefore, option (D) is correct.

Note:
The nine-point circle of a triangle is a circle going through 9 special points:
1) Midpoints of three sides of the triangle.
2) Three foot of perpendicular from vertices to front side.
3) Three midpoints from the vertices to the orthocenter of the triangle.