Chapter 10.2, Problem 39E

To determine

To write: The equation of the circle who is tangent to the line with equation $x+3y=-2$ and has its center at $\left(5,1\right)$ .

Answer to Problem 39E

  ${\left(x-\frac{11}{2}\right)}^2+{\left(y-\frac{5}{2}\right)}^2=\frac{25}{2}$

Explanation of Solution

Given information: Circle who is tangent to the line with equation $x+3y=-2$ and has its center at $\left(5,1\right)$

Calculation:

If we have the center then we have so far:

  ${\left(x-5\right)}^2+{\left(y-1\right)}^2=r^2$

If the circle is “tangent” to a line, then they meet at a point. At that point they are equal. So we set the equations equal to each other.

  ${\left(x-5\right)}^2+{\left(y-1\right)}^2=x+3y+2$

Expand the terms and collect them.

  $\begin{array}{l}{x}^2-10x+25+y^2-2y+1=x+3y+2\\ x^2-11x+y^2-5y+24=0\end{array}$

Group terms then complete the square:

  $\left(x^2-11x--\right)+\left(y^2-5y--\right)=-24$

Add ${\left(\frac{1}{2}b\right)}^2$ from each term to both sides $b=\left(-11,-5\right)$ respectively

  $\left(x^2-11x+\frac{121}{4}\right)+\left(y^2-5y+\frac{25}{4}\right)=-24+\frac{121}{4}+\frac{25}{4}$

Factor and simplify

Final equation:

  ${\left(x-\frac{11}{2}\right)}^2+{\left(y-\frac{5}{2}\right)}^2=\frac{25}{2}$