Question
Mathematics Question on Equation of a Line in Space
The incentre of an equilateral triangle is (1,1) and the equation of one side is 3x+4y+3=0 Then, the equation of the circumcircle of the 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+14=0
Answer
x2+y2−2x−2y−14=0
Explanation
Solution
Since, triangle is equilateral therefore incentre (1, 1) lies on the centroid of the △ABC.
∴GD= Length of perpendicular from the point
G(1,1) to the line 3x+4y+3=0
3x+4y+3=0
=32+423(1)+4(1)+3=2
AG=2GD=4
∴ Equation of circumcircle with centre at (1,1) and radius =4 units
(x−1)2+(y−1)2=42
⇒x2+y2−2x−2y−14=0