Question

Equation of incircle of an equilateral triangle ABC where B≡(2,0) C≡(4,0) and A lies in the fourth quadrant is
A. $ {x^2} + {y^2} - 6x + \dfrac{{2y}}{{\sqrt 3 }} + 9 = 0 $
B. $ {x^2} + {y^2} - 6x - \dfrac{{2y}}{{\sqrt 3 }} + 9 = 0 $
C. $ {x^2} + {y^2} + 6x + \dfrac{{2y}}{{\sqrt 3 }} + 9 = 0 $
D.None of these

Answer 1

Hint: As ABC is an equilateral triangle, we can find the coordinates of A using the distance formula then draw the incircle with the help of angle bisectors. Further, find the radius of the circle and then apply the formula for finding the equation of the circle.

Complete step-by-step answer:
First, let us plot the given points on the graph,

Now on drawing a perpendicular from A to the x-axis, we see that point D is the midpoint of line BC as ABC is an equilateral triangle.

So the x coordinate of A will be the same as that of D, thus coordinates of point A can be written as (3,y).
As ABC is an equilateral triangle, length of BA= length of BC
Now length of BC= 4-2= 2 units
Applying the distance formula on BA, we get the length of BA $ = \sqrt {{{(3 - 2)}^2} + {{(y - 0)}^2}} = \sqrt {1 + {y^2}} $
Therefore,
  $
  \sqrt[{}]{{1 + {y^2}}} = 2 \\
  1 + {y^2} = 4 \\
  {y^2} = 4 - 1 = 3 \\
  y = \pm \sqrt 3 \;
  $
But point A lies in the 4th quadrant so $ y = \sqrt 3 $ is rejected,
Thus coordinates of point A is $ (3, - \sqrt 3 ) $
Now to find the incenter ( center of the incircle) we make angular bisectors of $ \angle B $ and $ \angle C $ . The point of intersection of these angular bisectors gives us the incenter of the incircle at point E.

As ABC is an equilateral triangle, $ \angle A = \angle B = \angle C = 60^\circ $
The angle bisector divides an angle into two equal parts, so $ \angle DBE = \angle DCE = 30^\circ $
And AD is perpendicular to BC so $ \angle EDB = \angle EDC = 90^\circ $
In triangle EDB, we see that
 

 $
  \tan 30 = \dfrac{{ED}}{{BD}} \\
  \dfrac{1}{{\sqrt 3 }} = \dfrac{{ED}}{1} \\
  ED = \dfrac{1}{{\sqrt 3 }} \;
  $
(BD=1 because D is the midpoint of BC)
i.e. distance of E from D is $ \dfrac{1}{{\sqrt 3 }} $ but point E is in 4th quadrant so its coordinates will be $ (3,\dfrac{{ - 1}}{{\sqrt 3 }}) $ .
ED is the radius of the incircle and E is the center of the incircle. Therefore we can find the equation of the circle by using this formula.
 $ {(x - a)^2} + {(y - b)^2} = {r^2} $
Where $ a = 3 $ , $ b = \dfrac{{ - 1}}{{\sqrt 3 }} $ and $ r = \dfrac{1}{{\sqrt 3 }} $
On solving the above equation, we get
 $
  {(x - 3)^2} + {(y - (\dfrac{{ - 1}}{{\sqrt 3 }}))^2} = {(\dfrac{1}{{\sqrt 3 }})^2} \\
  {x^2} + 9 - 6x + {(y + \dfrac{1}{{\sqrt 3 }})^2} = \dfrac{1}{3} \\
  {x^2} + 9 - 6x + {y^2} + \dfrac{1}{3} + \dfrac{{2y}}{{\sqrt 3 }} = \dfrac{1}{3} \\
  {x^2} + {y^2} - 6x + \dfrac{{2y}}{{\sqrt 3 }} + 9 = 0 \;
  $
Hence, option (A) is the correct answer.
So, the correct answer is “Option A”.

Note: The distance measured from any point on the y-axis, parallel to the x-axis, the x-coordinate is called abscissa. The distance measure from any point on the x-axis, parallel to the y-axis, the y-coordinate is called ordinate.
The points D, E and A all lie on the same line that’s why they have the same x-coordinate that is 3. The point of intersection of the three altitudes of a triangle is called the orthocentre of the triangle but in this question, we are talking about the incenter so don’t get confused between the two.