Question

Find the equation of the circumference inscribed to the triangle of sides: y=0, 3x-4y=0, and 4x+3y-50=0

Answer 1

`(x-6.25)^(2)+y^(2)=6.25^(2)`

Explanation:

Given the equations of the lines that represent the sides of the triangle, the diagram on the coordinate plane;

Given the lines and the three vertices of the triangle. Let P(a,b) be the center of the circle.

Therefore,

`PA^2=PB^2=PC^2`

If A(12.5,0), B(8,6) and C(0,0)

`(a-12.5)^2+(b-0)^2=(a-8)^2+(b-6)^2=(a-0)^2+(b-0)^2`

Solve for ''a'' and ''b'':

`\begin{gathered} a^2-25a+156.25+b^2=a^2+b^2 \\ -25a+156.25=0 \\ a=(-156.25)/(-25) \\ a=(25)/(4)=6.25 \\ \\ a^2-16a+64+b^2-12b+36=a^2-25a+156.25+b^2 \\ -16a+64-12b+36=-25a+156.25 \\ 9a-12b=156.25-64-36 \\ 9(6.25)-12b=56.25 \\ 12b=56.25-56.25 \\ b=0 \end{gathered}`

Then, the radius of the circle is:

`\begin{gathered} PA=\sqrt[2]{(a-12.5)^2+b^2} \\ PA=6.25 \end{gathered}`

Hence, the equation of the circle is given as:

`(x-6.25)^2+y^2=6.25^2`