Question

Two vertices of an equilateral triangle are (−1,0) and (1,0) and its third vertex lies above the x−axis. The equation of its circumcircle is________

Answer 1

HINT- Proceed the solution of this question by finding the first third vertex then we have coordinates of three points hence with the help of the general equation of the circle we can find the equation of the required circumcircle.

Complete step by step answer:
In this question it is given two verities of an equilateral triangle A (−1,0) and B (1,0)
Let AB be base of the Triangle
Hence if we draw the figure of triangle

From here we can find the value of h so that we can get coordinates of the third vertex.

As we know that each angle of equilateral triangle is ${60^0}$hence we can apply
$ \Rightarrow \tan {60^0} = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \dfrac{{\text{h}}}{1}$
$ \Rightarrow \sqrt 3 = \dfrac{{\text{h}}}{1} \Rightarrow {\text{h = 1}}$
Hence we can say the third point is at a height of \[\sqrt 3 {\text{ i}}{\text{.e}}{\text{. (0,}}\sqrt 3 {\text{)}}\] will be coordinate of third vertex.
General Equation of the circle is ${{\text{x}}^2} + {{\text{y}}^2} + 2{\text{gx + 2fy + c = 0}}$
Circle passes through A (−1,0)
Hence it will satisfy the equation of circle, so on putting Point A (−1,0) in general Equation of the circle
$ \Rightarrow {( - 1)^2} + {0^2} + 2{\text{g( - 1) + 2f}}{\text{.0 + c = 0}}$
⟹1−2g+c=0⋯ (1)
Circle passes through B (1,0), so similarly putting Point B (1,0) in general Equation of the circle
$ \Rightarrow {(1)^2} + {0^2} + 2{\text{g(1) + 2f}}{\text{.0 + c = 0}}$
⟹1+2g+c=0⋯ (2)
Circle also passes through C \[{\text{(0,}}\sqrt 3 {\text{)}}\]
$ \Rightarrow {0^2} + {\left( {\sqrt 3 } \right)^2} + 2{\text{g}} \times 0{\text{ + 2f}} \times \sqrt {\text{3}} {\text{ + c = 0}}$
$ \Rightarrow 3{\text{ + 2f}} \times \sqrt {\text{3}} {\text{ + c = 0}}$
⟹9+6​f+c=0⋯ (3)
From equation (1), (2), (3), solve for g, h and c
 Adding equation (1) and (2)
⟹2+2c=0
⟹ c = -1
On putting the value of c in equation (1)
⟹1−2g+-1=0
⟹ g=0
On putting the value of C in equation (3)
$ \Rightarrow 3{\text{ + 2f}} \times \sqrt {\text{3}} {\text{ - 1 = 0}}$
$ \Rightarrow {\text{f = }}\dfrac{{{\text{ - 2}}}}{{2\sqrt 3 }} = \dfrac{{ - 1}}{{\sqrt 3 }}$
⟹ C = −1, g = 0, ${\text{f = }}\dfrac{{ - 1}}{{\sqrt 3 }}$

Hence on putting g, f, c in general Equation of the circle is General Equation of the circle is ${{\text{x}}^2} + {{\text{y}}^2} + 2{\text{gx + 2fy + c = 0}}$
 $ \Rightarrow {{\text{x}}^2} + {{\text{y}}^2} + 2(0){\text{x + 2(}}\dfrac{{ - 1}}{{\sqrt 3 }}{\text{)y - 1 = 0}}$
$ \Rightarrow \sqrt 3 \left( {{{\text{x}}^2} + {{\text{y}}^2}} \right){\text{ - 2y - }}\sqrt 3 {\text{ = 0}}$

Therefore The equation of its circumcircle is $\sqrt 3 \left( {{{\text{x}}^2} + {{\text{y}}^2}} \right){\text{ - 2y - }}\sqrt 3 {\text{ = 0}}$ .

Note- In this particular question we should know that the circumcircle is a triangle's circumscribed circle, i.e., the unique circle which passes through each of the triangle's three vertices A ,B and C of the triangle. The center O of the circumcircle is called the circumcentre of that triangle, and the circle's radius R is called the circumradius. A triangle's three perpendicular bisectors P,Q and R meet at O as shown in the figure below.