Q.464. To find the equation of the in-circle given vertices of a triangle

Question 464.
Find the equation of circle inscribed in the triangle with vertices
(-7,-10),(-7,15) and (5,-1).

Answer 464.
Let A (-7, -10), B (-7, 15) and C (5, -1) be the given vertices of ΔABC

=> a = BC = √[(5+7)^2 + (-1-15)^2] = 20,
b = CA = √[(5+7)^2 + (-1+10)^2] = 15
and c = AB = √(-7+7)^2 + (15+10)^2] = 25

=> coordinates of the in-center are
x = 1/(a+b+c) * (-7a -7b+5c)
= 1/(20+15+25) * (-7*20 - 7*15 + 5*25) = - 2
and y = 1/(a+b+c) * (-10a +15b-c)
= 1/(20+15+25) * (-10*20 +15*15 - 25) = 0
=> incenter = (-2, 0)

Radius of the in-circle
= perpendicular distance from (-2, 0) to the line through AB
= -2 + 7 = 5

=> eqn. of the inscribed circle is
(x + 2)^2 + y^2 = 5^2
=> x^2 + y^2 + 4x - 21 = 0.

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