JEE Main 2015 · 11 Apr · Shift 1 · Mathematics Q79

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MathematicsSingle correct+4 / -1Question 79 of 90

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 :

x^2+y^2-2x-2y-2=0

x^2+y^2-2x-2y-14=0

x^2+y^2-2x-2y+2=0

x^2+y^2-2x-2y-7=0

Answer and worked solution

Official answer: B

The inradius is $\dfrac{|3+4+3|}5=2$ . For an equilateral triangle the circumradius is twice that, 4, with the same centre:
$(x-1)^2+(y-1)^2=16\iff x^2+y^2-2x-2y-14=0.$

Topics:Equation of a circle Distance of a point from a line

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JEE Main 2015 · 11 Apr · Shift 1, Mathematics question 79