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

Given Information:

• Incentre of equilateral triangle: (1, 1)
• Equation of one side: 3x + 4y + 3 = 0

For an equilateral triangle, the incentre, centroid, and circumcentre all coincide at the same point.

Therefore, the circumcentre is at (1, 1).

Finding the Circumradius:

The circumradius R is the perpendicular distance from the circumcentre (1, 1) to any side of the triangle.

Using the perpendicular distance formula from point (x₁, y₁) to line ax + by + c = 0:

d=∣ax1+by1+c∣ √a²+b² 

For the line 3x + 4y + 3 = 0 and point (1, 1):

R= ∣3(1)+4(1)+3∣ √3²+4²  

R= |3 + 4 + 3| √9+16 

R = 10/√25=10/5=2

Equation of Circumcircle:

The general equation of a circle with centre (h, k) and radius r is:

(x−h)² + (y−k)² = r²

With centre (1, 1) and radius 2:

(x−1)² +(y−1)² = 4 

Expanding:

x²− 2x + 1 + y²− 2y + 1 = 4 

x² + y² − 2x − 2y + 2 = 4 

x² + y² −2x−2y−2=0

Answer: Option (a) x² + y² - 2x - 2y - 2 = 0

Now, $r=\frac{|3+4+3|}{\sqrt{9+16}}=2$ 

$\Rightarrow R=4$

$\therefore$ Equation of circumcircle is $(x-1{)}^2+(y-1{)}^2=16$

$\Rightarrow x^2+y^2-2x-2y-14=0$