Question

(6,0), (0,6) and (7,7) are the vertices of a triangle. The circle inscribed in the triangle has equation
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) ${x^2} + {y^2} - 9x - 9y - 36 = 0$

Answer 1

Hint:
Let the triangle be ABC which has vertices A(6,0), B(0,6) and C(7,7) and AB, BC and AC are the sides with distances a, b and c respectively.
Using the distance formula, find the distances a, b and c.
Now, using the formula $O = \left( {\dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)$ , find the in-centre i.e. (h,k) of the circle.
Also, in-radius can be found using $r = \dfrac{{\left| {x + y - 6} \right|}}{{\sqrt {{a^2} + {b^2}} }}$ , where \[x + y = 6\] is the equation of line join point A and B.
Thus, the equation of circle can be found by ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$.

Complete step by step solution:
Let the triangle be ABC which has vertices A(6,0), B(0,6) and C(7,7) and AB, BC and AC are the sides with distances a, b and c respectively.
Using the distance formula, we can find the distances a, b and c
 $\Rightarrow a = \sqrt {{{\left( {7 - 0} \right)}^2} + {{\left( {7 - 6} \right)}^2}} $
        $
   = \sqrt {{{\left( 7 \right)}^2} + {{\left( 1 \right)}^2}} \\
   = \sqrt {49 + 1} \\
   = \sqrt {50} \\
   = 5\sqrt 2 \\
 $
 $\Rightarrow b = \sqrt {{{\left( {7 - 6} \right)}^2} + {{\left( {7 - 0} \right)}^2}} $
        $
   = \sqrt {{{\left( 1 \right)}^2} + {{\left( 7 \right)}^2}} \\
   = \sqrt {1 + 49} \\
   = \sqrt {50} \\
   = 5\sqrt 2 \\
 $
 $\Rightarrow c = \sqrt {{{\left( {6 - 0} \right)}^2} + {{\left( {0 - 6} \right)}^2}} $
         $
   = \sqrt {{{\left( 6 \right)}^2} + {{\left( { - 6} \right)}^2}} \\
   = \sqrt {36 + 36} \\
   = \sqrt {36\left( 2 \right)} \\
   = 6\sqrt 2 \\
 $
Let us assume the in-centre of the circle as O.
So, we get
  $
  O = \left( {\dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right) \\
   = \left( {\dfrac{{5\sqrt 2 \left( 6 \right) + 5\sqrt 2 \left( 0 \right) + 6\sqrt 2 \left( 7 \right)}}{{5\sqrt 2 + 5\sqrt 2 + 6\sqrt 2 }},\dfrac{{5\sqrt 2 \left( 0 \right) + 5\sqrt 2 \left( 6 \right) + 6\sqrt 2 \left( 7 \right)}}{{5\sqrt 2 + 5\sqrt 2 + 6\sqrt 2 }}} \right) \\
   = \left( {\dfrac{9}{2},\dfrac{9}{2}} \right) \\
 $
Also, the in-radius r of the in-circle is equal to the perpendicular distance from in-centre to any of the sides.
So, the equation of the line AB using two-point form can be formed as
 $
  y - 0 = \left( {\dfrac{{6 - 0}}{{0 - 6}}} \right)\left( {x - 6} \right) \\
  \Rightarrow y = \dfrac{6}{{ - 6}}\left( {x - 6} \right) \\
  \Rightarrow y = - \left( {x - 6} \right) \\
  \Rightarrow x + y - 6 = 0 \\
 $
Thus, the in-radius of the in-circle can be given by $r = \dfrac{{\left| {x + y - 6} \right|}}{{\sqrt {{a^2} + {b^2}} }}$ .
 $
  \Rightarrow r = \dfrac{{\left| {\dfrac{9}{2} + \dfrac{9}{2} - 6} \right|}}{{\sqrt {{1^2} + {1^2}} }} \\
  \Rightarrow r = \dfrac{{\left| {9 - 6} \right|}}{{\sqrt {1 + 1} }} \\
  \Rightarrow r = \dfrac{{\left| 3 \right|}}{{\sqrt 2 }} \\
 $
Thus, we get the radius of circle $r = \dfrac{3}{{\sqrt 2 }}$ .
The equation of the circle can be formed using the formula ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ , where $h = \dfrac{9}{2},k = \dfrac{9}{2},r = \dfrac{3}{{\sqrt 2 }}$ .
 \[
  \Rightarrow {\left( {x - \dfrac{9}{2}} \right)^2} + {\left( {y - \dfrac{9}{2}} \right)^2} = {\left( {\dfrac{3}{{\sqrt 2 }}} \right)^2} \\
  \Rightarrow {x^2} - 9x + \dfrac{{81}}{4} + {y^2} - 9y + \dfrac{{81}}{4} = \dfrac{9}{2} \\
  \Rightarrow {x^2} + {y^2} - 9x - 9y + 2\left( {\dfrac{{81}}{4}} \right) - \dfrac{9}{2} = 0 \\
  \Rightarrow {x^2} + {y^2} - 9x - 9y + \dfrac{{81}}{2} - \dfrac{9}{2} = 0 \\
  \Rightarrow {x^2} + {y^2} - 9x - 9y + \dfrac{{81 - 9}}{2} = 0 \\
  \Rightarrow {x^2} + {y^2} - 9x - 9y + \dfrac{{72}}{2} = 0 \\
  \Rightarrow {x^2} + {y^2} - 9x - 9y + 36 = 0 \\
 \]
Thus, the equation of the circle is \[{x^2} + {y^2} - 9x - 9y + 36 = 0\] .

Option (B) is correct.

Note:
Alternate Method:
Let the triangle be ABC which has vertices A(6,0), B(0,6) and C(7,7) and AB, BC and AC are the sides with distances a, b and c respectively.
Using the distance formula, we can find the distances a, b and c
 $\Rightarrow a = \sqrt {{{\left( {7 - 0} \right)}^2} + {{\left( {7 - 6} \right)}^2}} $
        $
   = \sqrt {{{\left( 7 \right)}^2} + {{\left( 1 \right)}^2}} \\
   = \sqrt {49 + 1} \\
   = \sqrt {50} \\
   = 5\sqrt 2 \\
 $
 $\Rightarrow b = \sqrt {{{\left( {7 - 6} \right)}^2} + {{\left( {7 - 0} \right)}^2}} $
        $
   = \sqrt {{{\left( 1 \right)}^2} + {{\left( 7 \right)}^2}} \\
   = \sqrt {1 + 49} \\
   = \sqrt {50} \\
   = 5\sqrt 2 \\
 $
 $\Rightarrow c = \sqrt {{{\left( {6 - 0} \right)}^2} + {{\left( {0 - 6} \right)}^2}} $
         $
   = \sqrt {{{\left( 6 \right)}^2} + {{\left( { - 6} \right)}^2}} \\
   = \sqrt {36 + 36} \\
   = \sqrt {36\left( 2 \right)} \\
   = 6\sqrt 2 \\
 $
Let us assume the in-centre of the circle as O.
So, we get
  $
  O = \left( {\dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right) \\
   = \left( {\dfrac{{5\sqrt 2 \left( 6 \right) + 5\sqrt 2 \left( 0 \right) + 6\sqrt 2 \left( 7 \right)}}{{5\sqrt 2 + 5\sqrt 2 + 6\sqrt 2 }},\dfrac{{5\sqrt 2 \left( 0 \right) + 5\sqrt 2 \left( 6 \right) + 6\sqrt 2 \left( 7 \right)}}{{5\sqrt 2 + 5\sqrt 2 + 6\sqrt 2 }}} \right) \\
   = \left( {\dfrac{9}{2},\dfrac{9}{2}} \right) \\
 $
Let M be the midpoint of the line AB.
$\Rightarrow $ The coordinates of M will be $\left( {\dfrac{{6 + 0}}{2},\dfrac{{0 + 6}}{2}} \right) = \left( {3,3} \right)$ .
Now, for radius we will find the distance OM using the distance formula between the points O $\left( {\dfrac{9}{2},\dfrac{9}{2}} \right)$ and M(3,3).
 $
  \left| {OM} \right| = \sqrt {{{\left( {\dfrac{9}{2} - 3} \right)}^2} + {{\left( {\dfrac{9}{2} - 3} \right)}^2}} \\
   = \sqrt {{{\left( {\dfrac{3}{2}} \right)}^2} + {{\left( {\dfrac{3}{2}} \right)}^2}} \\
   = \sqrt {{{\left( {\dfrac{3}{2}} \right)}^2}\left( {1 + 1} \right)} \\
   = \dfrac{3}{2}\sqrt 2 \\
   = \dfrac{3}{{\sqrt 2 }} \\
 $
Thus, $r = \dfrac{3}{{\sqrt 2 }}$ .
The equation of the circle can be formed using the formula ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ , where $h = \dfrac{9}{2},k = \dfrac{9}{2},r = \dfrac{3}{{\sqrt 2 }}$ .
 \[
  \Rightarrow {\left( {x - \dfrac{9}{2}} \right)^2} + {\left( {y - \dfrac{9}{2}} \right)^2} = {\left( {\dfrac{3}{{\sqrt 2 }}} \right)^2} \\
  \Rightarrow {x^2} - 9x + \dfrac{{81}}{4} + {y^2} - 9y + \dfrac{{81}}{4} = \dfrac{9}{2} \\
  \Rightarrow {x^2} + {y^2} - 9x - 9y + 2\left( {\dfrac{{81}}{4}} \right) - \dfrac{9}{2} = 0 \\
  \Rightarrow {x^2} + {y^2} - 9x - 9y + \dfrac{{81}}{2} - \dfrac{9}{2} = 0 \\
  \Rightarrow {x^2} + {y^2} - 9x - 9y + \dfrac{{81 - 9}}{2} = 0 \\
  \Rightarrow {x^2} + {y^2} - 9x - 9y + \dfrac{{72}}{2} = 0 \\
  \Rightarrow {x^2} + {y^2} - 9x - 9y + 36 = 0 \\
 \]
Thus, the equation of the circle is \[{x^2} + {y^2} - 9x - 9y + 36 = 0\] .