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Question: The vertices of a triangle are \(\left( {1,\sqrt 3 } \right),\left( {2\cos \theta ,2\sin \theta } \r...

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

Explanation

Solution

Hint: Find the coordinates of circumcentre and centroid of the given triangle. Then use the relation 0G:GH=1:20G:GH = 1:2, where HH is circumcentre, where OO is orthocentre and where GG is centroid to find the coordinates of orthocentre. Write the locus in terms of xx and yy.

Complete step-by-step answer:
We are given the coordinates of triangle as (1,3),(2cos⁡θ,2sin⁡θ)\left( {1,\sqrt 3 } \right),\left( {2\cos \theta ,2\sin \theta } \right) and (2sin⁡θ,−2cos⁡θ)\left( {2\sin \theta , - 2\cos \theta } \right).
Also, each of the coordinates of the triangle satisfies the equation x2+y2=4{x^2} + {y^2} = 4.
The equation x2+y2=4{x^2} + {y^2} = 4 represents the equation of the circle.
On comparing the equation of circle, x2+y2=4{x^2} + {y^2} = 4with the general form of equation, (x−h)2+(y−k)2=1{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = 1 , where (h,k)\left( {h,k} \right) are the coordinates of centre and rris the radius of circle.
Thus, x2+y2=4{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(0,0)O\left( {0,0} \right)
We will find the centroid of triangle using the formula, (x1+x2+x33,y1+y2+y33)\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=(2cos⁡θ+2sin⁡θ+13,2sin⁡θ−2cos⁡θ+33)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=1:20G:GH = 1:2, where HH is circumcentre.
We will find the coordinates of orthocentre (h,k)\left( {h,k} \right), using the section formula, (m(x1)+n(x2)m+n,m(y1)+n(y2)m+n)\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)
2cos⁡θ+2sin⁡θ+13=1(h)+2(0)1+2=h3 ⇒h=2cos⁡θ+2sin⁡θ+1 (1)  \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) \\\
2sin⁡θ−2cos⁡θ+33=1(k)+2(0)1+2=k3 ⇒k=2sin⁡θ−2cos⁡θ+3 (2)  \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⁡θ\sin \theta
h+k=2cos⁡θ+2sin⁡θ+1+2sin⁡θ−2cos⁡θ+3 ⇒h+k=4sin⁡θ+1+3 ⇒sin⁡θ=h+k−1−34 (3)  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⁡θ\cos \theta
h−k=2cos⁡θ+2sin⁡θ+1−2sin⁡θ+2cos⁡θ−3 ⇒h−k=4cos⁡θ+1−3 ⇒cos⁡θ=h−k−1+34 (4)  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⁡2θ+cos⁡2θ=(h+k−1−34)2+(h−k−1+34)2 ⇒16=(h+k−1−3)2+(h−k−1+3)2 ⇒16=(h−1)2+(k−3)2+2(h−1)(k−3)+(h−1)2+(k−3)2−2(h−1)(k−3) ⇒16=2((h−1)2+(k−3)2) ⇒(h−1)2+(k−3)2=8  {\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 hh by xx and kk by yy to write the equation of locus of orthocentre.
(x−1)2+(y−3)2=8{\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=1:20G:GH = 1:2, where HH is circumcentre, where OO is orthocentre and where GG is centroid.