If the incentre of an equilateral triangle is (1, 1) and the equation of its one side is 3x + 4y + 3 = 0, then the equation of the circumcircle of this triangle is :

Question

If the incentre of an equilateral triangle is (1, 1) and the equation of its one side is 3x + 4y + 3 = 0, then the equation of the circumcircle of this triangle is : (A) x2+y2-2x-2y-2=0 (B) x2+y2-2x-2y-14=0 (C) x2+y2-2x-2y+2=0 (D) x2+y2-2x-2y-7=0

Tardigrade Admin · Accepted Answer

Let radius of circumcircle be According to the question, (r/2)=(10/5) ⇒ r=4 So equation of required circle is (x-1)2+(y-1)2=16 ⇒ x2+y2-2x-2y-14=0

Q. If the incentre of an equilateral triangle is (1, 1) and the equation of its one side is $3 x + 4 y + 3 = 0$ , then the equation of the circumcircle of this triangle is :

$x^{2} + y^{2} - 2 x - 2 y - 2 = 0$

$x^{2} + y^{2} - 2 x - 2 y - 14 = 0$

$x^{2} + y^{2} - 2 x - 2 y + 2 = 0$

$x^{2} + y^{2} - 2 x - 2 y - 7 = 0$

Let radius of circumcircle be According to the question,
$\frac{r}{2} = \frac{10}{5} \Rightarrow r = 4$
So equation of required circle is
$(x - 1)^{2} + (y - 1)^{2} = 16$
$\Rightarrow x^{2} + y^{2} - 2 x - 2 y - 14 = 0$

Q. If the incentre of an equilateral triangle is (1, 1) and the equation of its one side is 3x+4y+3=0, then the equation of the circumcircle of this triangle is :

x2+y2−2x−2y−2=0

x2+y2−2x−2y−14=0

x2+y2−2x−2y+2=0

x2+y2−2x−2y−7=0