`(6, 0), (0, 6) and (7, 7)` are the vertices of a triangle. The circle inscribed in the triangle has the equation

Options

A. `x^(2)+y^(2)-9x-9y+36=0`
B. `x^(2)+y^(2)+9x-9y+36=0`
C. `x^(2)+y^(2)+9x+9y-36=0`
D. `x^(2)+y^(2)+18x-18y+36=0`

Answer

2

Explanation

The triangle is evidently isosceles and, therefore, the median through C is the angle bisector of `/_C`.
The equation of the angle bisector is `y =- x` and the incenter
`I-=(-a,a)` where a is positive.
The equation of AC is `y-0= -7(x+6), i.e., 7x+y+42=0`
and the equation of AB is `x-y+6=0`.
The lengths of the perpendiculars from I to AB and AC are equal. Therefore,
`|(-7a=a+42)/(sqrt(50))|=|(-a-a+6)/(sqrt(2))|`
or `a=(9)/(2)` `( :' agt0)`
Therefore, the center is `(-9//2, 9//2)` and the radius is `3//sqrt(2)`.
Hence, the equation of the circle is
`(x+(9)/(2))^(2)+(y-(9)/(2))^(2)=(9)/(2)`
or `x^(2)+y^(2)+9x-9y+36=0`