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Question
The equation of the circucircle of an equilateral triangle is x2 + y2 + 2gx + 2fy + c = 0 and one vertex of the triangle is (1, 1). Find the equation of the incircle of the triangle
The equation of the circucircle of an equilateral triangle is x2 + y2 + 2gx + 2fy + c = 0 and one vertex of the triangle is (1, 1). Find the equation of the incircle of the triangle
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Solution
x2 + y2 + 2gx + 2fy + c = 0
So the equation of the circle in the standard form is (x-g' )2 + (y-f')2 = a2 ( with centre (g',f') and radius a)
So x2 + y2 + 2gx + 2fy + c = 0 Or (x +g)2 + (y +f)2 = -c +g2 + f2
Here a is the radius and the distance between the centre and the vertex, so a2 = (-g-1)2 + (-f-1)2
So the centre of the circle is (-g,-f) and the radius = √((-g-1)2 + (-f-1)2)
In an equilateral triangle, the circumcentre and incentre coincides, and the radius of the circumcircle is 2 times the radius of the incircle.
Hence the incentre is (-g,-f) and the radius would be √((-g-1)2 + (-f-1)2 )/2
So the equation of the incircle is (x + g)2 + (y +f)2 = ((-g-1)2 + (-f-1)2 )/4
So the equation of the circle in the standard form is (x-g' )2 + (y-f')2 = a2 ( with centre (g',f') and radius a)
So x2 + y2 + 2gx + 2fy + c = 0 Or (x +g)2 + (y +f)2 = -c +g2 + f2
Here a is the radius and the distance between the centre and the vertex, so a2 = (-g-1)2 + (-f-1)2
So the centre of the circle is (-g,-f) and the radius = √((-g-1)2 + (-f-1)2)
In an equilateral triangle, the circumcentre and incentre coincides, and the radius of the circumcircle is 2 times the radius of the incircle.
Hence the incentre is (-g,-f) and the radius would be √((-g-1)2 + (-f-1)2 )/2
So the equation of the incircle is (x + g)2 + (y +f)2 = ((-g-1)2 + (-f-1)2 )/4
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