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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?

x2+y2-2x-2y-14=0

x2+y2-2x-2y-14=0

x2+y2-2x-2y-14=0

x2+y2-2x-2y+2=0

Solution:

r(inradius)=|3(1)+4(1)+3|/√3²+4²=2
As for equilateral triangle circumcentre and incentre coincide.
sin30=r/R
R=4
Equation of circle with centre(1,1) and radius=4
(x-1)²+(y-1)²=4²=16
x²+y²-2x-2y-14=0