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Question

Mathematics Question on Equation of a Line in Space

The incentre of an equilateral triangle is (1,1)(1,1) and the equation of one side is 3x+4y+3=03 x+4 y+3=0 Then, the equation of the circumcircle of the triangle is

A

x2+y22x2y2=0x^{2}+y^{2}-2x-2y-2=0

B

x2+y22x2y14=0x^{2}+y^{2}-2x-2y-14=0

C

x2+y22x2y+2=0x^{2}+y^{2}-2x-2y+2=0

D

x2+y22x2y+14=0x^{2}+y^{2}-2x-2y+14=0

Answer

x2+y22x2y14=0x^{2}+y^{2}-2x-2y-14=0

Explanation

Solution

Since, triangle is equilateral therefore incentre (1, 1) lies on the centroid of the ABC\triangle A B C.
GD=\therefore G D= Length of perpendicular from the point
G(1,1)G(1,1) to the line 3x+4y+3=03 x+4 y+3=0

3x+4y+3=03 x+4 y+3=0
=3(1)+4(1)+332+42=2=\frac{3(1)+4(1)+3}{\sqrt{3^{2}+4^{2}}}=2
AG=2GD=4A G=2 G D=4
\therefore Equation of circumcircle with centre at (1,1)(1,1) and radius =4=4 units
(x1)2+(y1)2=42(x-1)^{2}+(y-1)^{2} =4^{2}
x2+y22x2y14=0\Rightarrow x^{2}+ y^{2}-2 x-2 y-14 =0