Question

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

A
x2+y29x+9y+36=0
B
x2+y29x9y+36=0
C
x2+y2+9x9y+36=0
D
x2+y29x9y36=0

Solution

## The correct option is B $$x^2+y^2-9x-9y+36=0$$Let, $$\triangle ABC$$ has the vertices $$A(6,0), B(0,6), C(7,7)$$ and $$AB, BC, CA$$ are the sides with distances $$c, a, b$$ respectively.$$\therefore$$ Using distance formula, $$a=\left| BC \right| =\sqrt { { (7-0) }^{ 2 }+{ (7-6) }^{ 2 } } =5\sqrt { 2 } \\ b=\left| CA \right| =\sqrt { { (7-6) }^{ 2 }+{ (7-0) }^{ 2 } } =5\sqrt { 2 } \\ c=\left| AB \right| =\sqrt { { (6-0) }^{ 2 }+{ (0-6) }^{ 2 } } =6\sqrt { 2 }$$Now, assuming the In-centre of the circle as $$O$$ we get,                          $$O=\left( \dfrac { a{ x }_{ 1 }+b{ x }_{ 2 }+c{ x }_{ 3 } }{ a+b+c } ,\dfrac { a{ y }_{ 1 }+b{ y }_{ 2 }+c{ y }_{ 3 } }{ a+b+c } \right) \\$$$$=\left( \dfrac { 5\sqrt { 2 } (6)+5\sqrt { 2 } (0)+6\sqrt { 2 } (7) }{ 5\sqrt { 2 } +5\sqrt { 2 } +6\sqrt { 2 } } ,\dfrac { 5\sqrt { 2 } (0)+5\sqrt { 2 } (6)+6\sqrt { 2 } (7) }{ 5\sqrt { 2 } +5\sqrt { 2 } +6\sqrt { 2 } } \right) \\$$$$\Rightarrow O=\left( \dfrac { 9 }{ 2 } ,\dfrac { 9 }{ 2 } \right)$$ and,             The inradius r is equal to the perpendicular distance from incenter to any of the sidesEquation of the line $$AB$$$$y-0=\dfrac{6-0}{0-6}(x-6)$$   $$[two-point\ form]$$$$y=\dfrac{6}{-6}(x-6)$$$$y=-(x-6)$$$$x+y-6=0$$So the inradius is$$r=\dfrac{|\dfrac{9}{2}+\dfrac{9}{2}-6|}{\sqrt{1+1}}$$-$$r=\dfrac{|9-6|}{\sqrt{2}}$$$$r=\dfrac{|3|}{\sqrt{2}}$$$$\therefore\ r=\dfrac{|3|}{\sqrt{2}}$$The equation of the incircle is$$(x-\dfrac{9}{2})^2+(y-\dfrac{9}{2})^2=\left(\dfrac{|3|}{\sqrt{2}} \right)^2$$$$x^2+\dfrac{81}{4}-9x+y^2+\dfrac{81}{4}-9y=\dfrac{9}{2}$$$$x^2+-9x+y^2-9y+\dfrac{81}{4}+\dfrac{81}{4}-\dfrac{9}{2}=0$$$$x^2+y^2-9x-9y+\dfrac{81}{2}-\dfrac{9}{2}=0$$$$x^2+y^2-9x-9y+\dfrac{81-9}{2}=0$$$$x^2+y^2-9x-9y+\dfrac{72}{2}=0$$$$x^2+y^2-9x-9y+36=0$$Maths

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