Question

The vertices of a triangle are $\left( {1,\sqrt 3 } \right),\left( {2\cos \theta ,2\sin \theta } \right)$ and $\left( {2\sin \theta , - 2\cos \theta } \right)$ where $\theta \in R$. The locus of orthocentre of the triangle is
A.${\left( {x - 1} \right)^2} + {\left( {y - \sqrt 3 } \right)^2} = 4$
B.${\left( {x - 2} \right)^2} + {\left( {y - \sqrt 3 } \right)^2} = 4$
C.${\left( {x - 1} \right)^2} + {\left( {y - \sqrt 3 } \right)^2} = 8$
D.${\left( {x - 2} \right)^2} + {\left( {y - \sqrt 3 } \right)^2} = 8$

Answer 1

Hint: Find the coordinates of circumcentre and centroid of the given triangle. Then use the relation $0G:GH = 1:2$, where $H$ is circumcentre, where $O$ is orthocentre and where $G$ is centroid to find the coordinates of orthocentre. Write the locus in terms of $x$ and $y$.

Complete step-by-step answer:
We are given the coordinates of triangle as $\left( {1,\sqrt 3 } \right),\left( {2\cos \theta ,2\sin \theta } \right)$ and $\left( {2\sin \theta , - 2\cos \theta } \right)$.
Also, each of the coordinates of the triangle satisfies the equation ${x^2} + {y^2} = 4$.
The equation ${x^2} + {y^2} = 4$ represents the equation of the circle.
On comparing the equation of circle, ${x^2} + {y^2} = 4$with the general form of equation, ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = 1$ , where $\left( {h,k} \right)$ are the coordinates of centre and $r$is the radius of circle.
Thus, ${x^2} + {y^2} = 4$ represents the equation of circle with origin as centre and radius of 2 units.
Hence, the circumcentre of the triangle is at origin. That is, circumcentre is $O\left( {0,0} \right)$
We will find the centroid of triangle using the formula, $\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$
Hence, centroid for the given triangle is $G = \left( {\dfrac{{2\cos \theta + 2\sin \theta + 1}}{3},\dfrac{{2\sin \theta - 2\cos \theta + \sqrt 3 }}{3}} \right)$
The relation between circumcentre, orthocentre and centroid of the triangle is given as $0G:GH = 1:2$, where $H$ is circumcentre.
We will find the coordinates of orthocentre $\left( {h,k} \right)$, using the section formula, \[\left( {\dfrac{{m\left( {{x_1}} \right) + n\left( {{x_2}} \right)}}{{m + n}},\dfrac{{m\left( {{y_1}} \right) + n\left( {{y_2}} \right)}}{{m + n}}} \right)\]
$
  \dfrac{{2\cos \theta + 2\sin \theta + 1}}{3} = \dfrac{{1\left( h \right) + 2\left( 0 \right)}}{{1 + 2}} = \dfrac{h}{3} \\
   \Rightarrow h = 2\cos \theta + 2\sin \theta + 1{\text{ }}\left( 1 \right) \\
$
$
  \dfrac{{2\sin \theta - 2\cos \theta + \sqrt 3 }}{3} = \dfrac{{1\left( k \right) + 2\left( 0 \right)}}{{1 + 2}} = \dfrac{k}{3} \\
   \Rightarrow k = 2\sin \theta - 2\cos \theta + \sqrt 3 {\text{ }}\left( 2 \right) \\
 $
Add equation (1) and (2), to get the value of $\sin \theta $
$
  h + k = 2\cos \theta + 2\sin \theta + 1 + 2\sin \theta - 2\cos \theta + \sqrt 3 \\
   \Rightarrow h + k = 4\sin \theta + 1 + \sqrt 3 \\
   \Rightarrow \sin \theta = \dfrac{{h + k - 1 - \sqrt 3 }}{4}{\text{ }}\left( 3 \right) \\
$
Similarly, subtract equation (1) and (2), to get the value of $\cos \theta $
$
  h - k = 2\cos \theta + 2\sin \theta + 1 - 2\sin \theta + 2\cos \theta - \sqrt 3 \\
   \Rightarrow h - k = 4\cos \theta + 1 - \sqrt 3 \\
   \Rightarrow \cos \theta = \dfrac{{h - k - 1 + \sqrt 3 }}{4}{\text{ }}\left( 4 \right) \\
$
Squaring and adding equation (3) and (4).
$
  {\sin ^2}\theta + {\cos ^2}\theta = {\left( {\dfrac{{h + k - 1 - \sqrt 3 }}{4}} \right)^2} + {\left( {\dfrac{{h - k - 1 + \sqrt 3 }}{4}} \right)^2} \\
   \Rightarrow 16 = {\left( {h + k - 1 - \sqrt 3 } \right)^2} + {\left( {h - k - 1 + \sqrt 3 } \right)^2} \\
   \Rightarrow 16 = {\left( {h - 1} \right)^2} + {\left( {k - \sqrt 3 } \right)^2} + 2\left( {h - 1} \right)\left( {k - \sqrt 3 } \right) + {\left( {h - 1} \right)^2} + {\left( {k - \sqrt 3 } \right)^2} - 2\left( {h - 1} \right)\left( {k - \sqrt 3 } \right) \\
   \Rightarrow 16 = 2\left( {{{\left( {h - 1} \right)}^2} + {{\left( {k - \sqrt 3 } \right)}^2}} \right) \\
   \Rightarrow {\left( {h - 1} \right)^2} + {\left( {k - \sqrt 3 } \right)^2} = 8 \\
$
Replace $h$ by $x$ and $k$ by $y$ to write the equation of locus of orthocentre.
\[{\left( {x - 1} \right)^2} + {\left( {y - \sqrt 3 } \right)^2} = 8\]
Hence, option C is correct.

Note: The orthocentre is the intersecting point of all altitudes of the triangle. The centroid in a triangle is the intersecting point of all the medians. The circumcentre is the intersection of all the perpendicular bisectors of a triangle. The relation between orthocentre, circumcentre and centroid is given by, $0G:GH = 1:2$, where $H$ is circumcentre, where $O$ is orthocentre and where $G$ is centroid.