Question

Let the orthocentre and centroid of the triangle \[A( - 3,5)\] and \[B(3,3)\] respectively. IF $C$ is the circumcenter of the triangle, then the radius of the circle having the segment $AC$ as diameter is ?

Answer 1

Hint: In a non-equilateral triangle, the circumference, the centroid and the orthocentre are collinear.Centroid (B) divides the line connecting orthocentre (A) and circumference(C) in the ratio \[2:1\]. Thus, using the Euler’s Line formula, we get, \[(\dfrac{{m{x_2} + n{x_1}}}{{m + n}} + \dfrac{{m{y_2} + n{y_1}}}{{m + n}})\].

Complete step by step answer:
Given data is as below,
Orthocentre point is \[A( - 3,5)\]= A(x1, y1).
Centroid point is \[B(3,3)\]= B(x2, y2).
Circumference point is not given.
Let the coordinates of circumference point C be (x, y).Thus, AB/ BC is in the ratio \[\dfrac{2}{1}\]= m/n. So, B divides AC in the ratio \[2:1\]= m:n. Thus, using the Euler’s Line formula, we get,
\[(\dfrac{{m{x_2} + n{x_1}}}{{m + n}} + \dfrac{{m{y_2} + n{y_1}}}{{m + n}})\]
Here, the given values are
\[m = 2,n = 1 \\
\Rightarrow {x_1} = - 3,{x_2} = x \\
\Rightarrow {y_1} = 5,{y_2} = y \\ \]
So, now we will solve the x coordinate part first and we have,
\[\dfrac{{m{x_2} + n{x_1}}}{{m + n}}\]
Substituting the values in the above equation, we get,
\[ \Rightarrow 3 = \dfrac{{2(x) + 1( - 3)}}{{2 + 1}}\]
Removing the brackets we get,
\[ \Rightarrow 3 = \dfrac{{2x - 3}}{3}\]
\[ \Rightarrow 9 = 2x - 3\]
Simplify the above expression, we get,
\[\Rightarrow 9 + 3 = 2x \\
\Rightarrow 12 = 2x \\
\Rightarrow 2x = 12 \\ \]
\[\Rightarrow x = \dfrac{{12}}{2} \\
\Rightarrow x = 6 \\ \]
Next, we will solve the y coordinate part and we have,
\[\dfrac{{m{y_2} + n{y_1}}}{{m + n}}\]
Substituting the values in the above formula, we get,
\[ \Rightarrow 3 = \dfrac{{2(y) + 1(5)}}{{2 + 1}}\]
Removing the brackets we get,
\[ \Rightarrow 3 = \dfrac{{2y + 5}}{3}\]
\[ \Rightarrow 9 = 2y + 5\]
Simplify the above expression we get,
\[\Rightarrow 9 - 5 = 2y \\
\Rightarrow 4 = 2y \\ \]
\[ \Rightarrow 2y = 4\]
\[\Rightarrow y = \dfrac{4}{2} \\
\Rightarrow y = 2 \]
Thus, the coordinates of C(x, y) =\[C(6,2)\]. Now, the diameter AC formula is where, \[A( - 3,5)\] and \[C(6,2)\].
\[\sqrt {{{(x - {x_1})}^2} + {{(y - {y_1})}^2}} \]
Substituting the values in the above equation, we get,
\[ \text{Diameter (AC)}= \sqrt {{{(6 - ( - 3))}^2} + {{(2 - 5)}^2}} \]
Removing the brackets we get,
\[\text{Diameter (AC)} = \sqrt {{{(6 + 3)}^2} + {{(2 - 5)}^2}} \]
\[\Rightarrow \text{Diameter (AC)} = \sqrt {{{(9)}^2} + {{( - 3)}^2}} \]
Squaring the numbers we get,
\[\text{Diameter (AC)}= \sqrt {81 + 9} \\
\Rightarrow \text{Diameter (AC)}= \sqrt {90} \\ \]
\[\Rightarrow \text{Diameter (AC)} = 3\sqrt {9 \times 10} \]
Reducing the given expression we get,
\[\text{Diameter (AC)}= 3\sqrt {10} \]
Since the radius is half the diameter.
So, Radius is = Diameter / 2
Radius =AC/2
\[\text{Radius}=\dfrac{{3\sqrt {10} }}{2}\]
\[\Rightarrow \text{Radius} = \dfrac{3}{2}\sqrt {10} \]
\[\Rightarrow \text{Radius}= 3 \times \sqrt {\dfrac{{10}}{4}} \]
\[\therefore \text{Radius} = 3 \times \sqrt {\dfrac{5}{2}} \]Units

Thus, the radius of the circle mentioned in the question is \[ = 3 \times \sqrt {\dfrac{5}{2}} \] units.

Note: We need to find the radius of the circle. Since the radius is half the diameter.Also we are using Euler’s line formula to find the coordinates of the given points. Please read the question carefully before answering the question, sometimes they might ask only diameter too and you will waste your time in finding the radius of the circle.