# (-2,1) and (4,1) are the endpoints of one chord of the circle, and (-2,-3) and (4,-3) are the endpoints of another chord of the circle. What is the center, radius, and equation?

Mar 28, 2017

Center is $\left(1 , - 1\right)$, radius is $\sqrt{13}$ and equation of circle is ${x}^{2} + {y}^{2} - 2 x + 2 y - 11 = 0$

#### Explanation:

As the ordinates of the endpoints of the chord joining $\left(- 2 , 1\right)$ and $\left(4 , 1\right)$ are equal, its perpendicular bisector is parallel to $y$-axis and its length is $4 - \left(- 2\right) = 6$. Note midpoint of chord is $\left(\frac{- 2 + 4}{2} , \frac{1 + 1}{2}\right)$ or $\left(1 , 1\right)$.

Similarly, as the ordinates of the endpoints of the chord joining $\left(- 2 , - 3\right)$ and $\left(4 , - 3\right)$ are equal, its perpendicular bisector is also parallel to $y$-axis and its length too is $4 - \left(- 2\right) = 6$. Note midpoint of chord is $\left(\frac{- 2 + 4}{2} , \frac{- 3 - 3}{2}\right)$ or $\left(1 , - 3\right)$

and hence two chords are equal and parallel and hence center is midpoint of the segment joining their midpoints i.e. $\left(\frac{1 + 1}{2} , \frac{1 - 3}{2}\right)$ or $\left(1 , - 1\right)$.`

Hence, radius is distance between $\left(1 , - 1\right)$ and say $\left(4 , 1\right)$ i.e.

$\sqrt{{\left(4 - 1\right)}^{2} + {\left(1 + 1\right)}^{2}} = \sqrt{13}$ and equation of circle is

${\left(x - 1\right)}^{2} + {\left(y + 1\right)}^{2} = 13$ i.e. ${x}^{2} + {y}^{2} - 2 x + 2 y - 11 = 0$

graph{x^2+y^2-2x+2y-11=0 [-8.92, 11.08, -6.32, 3.68]}