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Question: (6,0), (0,6) and (7,7) are the vertices of a triangle. The circle inscribed in the triangle has equa...

(6,0), (0,6) and (7,7) are the vertices of a triangle. The circle inscribed in the triangle has equation
A) x2+y2−9x+9y+36=0{x^2} + {y^2} - 9x + 9y + 36 = 0
B) x2+y2−9x−9y+36=0{x^2} + {y^2} - 9x - 9y + 36 = 0
C) x2+y2+9x−9y+36=0{x^2} + {y^2} + 9x - 9y + 36 = 0
D) x2+y2−9x−9y−36=0{x^2} + {y^2} - 9x - 9y - 36 = 0

Explanation

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, find the distances a, b and c.
Now, using the formula O=(ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c)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=∣x+y−6∣a2+b2r = \dfrac{{\left| {x + y - 6} \right|}}{{\sqrt {{a^2} + {b^2}} }} , where x+y=6x + y = 6 is the equation of line join point A and B.
Thus, the equation of circle can be found by (x−h)2+(y−k)2=r2{\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
⇒a=(7−0)2+(7−6)2\Rightarrow a = \sqrt {{{\left( {7 - 0} \right)}^2} + {{\left( {7 - 6} \right)}^2}}
=(7)2+(1)2 =49+1 =50 =52  = \sqrt {{{\left( 7 \right)}^2} + {{\left( 1 \right)}^2}} \\\ = \sqrt {49 + 1} \\\ = \sqrt {50} \\\ = 5\sqrt 2 \\\
⇒b=(7−6)2+(7−0)2\Rightarrow b = \sqrt {{{\left( {7 - 6} \right)}^2} + {{\left( {7 - 0} \right)}^2}}
=(1)2+(7)2 =1+49 =50 =52  = \sqrt {{{\left( 1 \right)}^2} + {{\left( 7 \right)}^2}} \\\ = \sqrt {1 + 49} \\\ = \sqrt {50} \\\ = 5\sqrt 2 \\\
⇒c=(6−0)2+(0−6)2\Rightarrow c = \sqrt {{{\left( {6 - 0} \right)}^2} + {{\left( {0 - 6} \right)}^2}}
=(6)2+(−6)2 =36+36 =36(2) =62  = \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=(ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c) =(52(6)+52(0)+62(7)52+52+62,52(0)+52(6)+62(7)52+52+62) =(92,92)  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=(6−00−6)(x−6) ⇒y=6−6(x−6) ⇒y=−(x−6) ⇒x+y−6=0  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=∣x+y−6∣a2+b2r = \dfrac{{\left| {x + y - 6} \right|}}{{\sqrt {{a^2} + {b^2}} }} .
⇒r=∣92+92−6∣12+12 ⇒r=∣9−6∣1+1 ⇒r=∣3∣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=32r = \dfrac{3}{{\sqrt 2 }} .
The equation of the circle can be formed using the formula (x−h)2+(y−k)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} , where h=92,k=92,r=32h = \dfrac{9}{2},k = \dfrac{9}{2},r = \dfrac{3}{{\sqrt 2 }} .

⇒(x−92)2+(y−92)2=(32)2 ⇒x2−9x+814+y2−9y+814=92 ⇒x2+y2−9x−9y+2(814)−92=0 ⇒x2+y2−9x−9y+812−92=0 ⇒x2+y2−9x−9y+81−92=0 ⇒x2+y2−9x−9y+722=0 ⇒x2+y2−9x−9y+36=0  \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 x2+y2−9x−9y+36=0{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
⇒a=(7−0)2+(7−6)2\Rightarrow a = \sqrt {{{\left( {7 - 0} \right)}^2} + {{\left( {7 - 6} \right)}^2}}
=(7)2+(1)2 =49+1 =50 =52  = \sqrt {{{\left( 7 \right)}^2} + {{\left( 1 \right)}^2}} \\\ = \sqrt {49 + 1} \\\ = \sqrt {50} \\\ = 5\sqrt 2 \\\
⇒b=(7−6)2+(7−0)2\Rightarrow b = \sqrt {{{\left( {7 - 6} \right)}^2} + {{\left( {7 - 0} \right)}^2}}
=(1)2+(7)2 =1+49 =50 =52  = \sqrt {{{\left( 1 \right)}^2} + {{\left( 7 \right)}^2}} \\\ = \sqrt {1 + 49} \\\ = \sqrt {50} \\\ = 5\sqrt 2 \\\
⇒c=(6−0)2+(0−6)2\Rightarrow c = \sqrt {{{\left( {6 - 0} \right)}^2} + {{\left( {0 - 6} \right)}^2}}
=(6)2+(−6)2 =36+36 =36(2) =62  = \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=(ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c) =(52(6)+52(0)+62(7)52+52+62,52(0)+52(6)+62(7)52+52+62) =(92,92)  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 (6+02,0+62)=(3,3)\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 (92,92)\left( {\dfrac{9}{2},\dfrac{9}{2}} \right) and M(3,3).
∣OM∣=(92−3)2+(92−3)2 =(32)2+(32)2 =(32)2(1+1) =322 =32  \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=32r = \dfrac{3}{{\sqrt 2 }} .
The equation of the circle can be formed using the formula (x−h)2+(y−k)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} , where h=92,k=92,r=32h = \dfrac{9}{2},k = \dfrac{9}{2},r = \dfrac{3}{{\sqrt 2 }} .

⇒(x−92)2+(y−92)2=(32)2 ⇒x2−9x+814+y2−9y+814=92 ⇒x2+y2−9x−9y+2(814)−92=0 ⇒x2+y2−9x−9y+812−92=0 ⇒x2+y2−9x−9y+81−92=0 ⇒x2+y2−9x−9y+722=0 ⇒x2+y2−9x−9y+36=0  \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 x2+y2−9x−9y+36=0{x^2} + {y^2} - 9x - 9y + 36 = 0 .