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The vertices of a triangle are the points of intersection of the line y = -x-5, the line x=2, and y = -1/5x+13/5. Find an equation of the circle passing through all three vertices.

May 1, 2022

1+0 Answers

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Pick a pair of equations and solve the system of those equations, and you will get the vertices.

$$\begin{cases}y = -x-5\\x=2\end{cases} \implies (x, y) = (2, -7)$$

$$\begin{cases}y = -x-5\\y = -\dfrac15x + \dfrac{13}5\end{cases}\implies (x, y) = \left(-\dfrac{19}2, \dfrac{9}2\right)$$

$$\begin{cases}y = -\dfrac15 x + \dfrac{13}5\\x=2\end{cases} \implies (x, y) = \left(2, \dfrac{11}5\right)$$

We find the center of the circle by solving the system

$$\begin{cases}(x - 2)^2 + (y + 7)^2 = \left(x + \dfrac{19}2\right)^2 + \left(y - \dfrac92\right)^2\\ \left(x + \dfrac{19}2\right)^2 + \left(y - \dfrac92\right)^2 = (x - 2)^2 + \left(y - \dfrac{11}5\right)^2\end{cases}$$

which just says the squared distance from the center to each vertex is the same.

Solving gives the center coordinates $$\left(-\dfrac{49}{10}, -\dfrac{12}5\right)$$.

Radius of circle is just the distance between $$\left(-\dfrac{49}{10}, -\dfrac{12}5\right)$$ and any one of the vertices, i.e., $$\sqrt{\left(-\dfrac{49}{10}-2\right)^2+\left(-\dfrac{12}5-\dfrac{11}5\right)^2} = \dfrac{23\sqrt{13}}{10}$$.

We use center-radius form of circle equation, and we have the answer: $$\left(x + \dfrac{49}{10}\right)^2 + \left(y + \dfrac{12}5\right)^2 = \dfrac{6877}{100}$$

If you want a simpler answer, multiply both sides by 100 and get $$(10x + 49)^2 + (10y + 24)^2 = 6877$$.

Exercise: Fill in the missed details in my solution.

May 2, 2022