# Points (–9, 2) and (–5, 6) are endpoints of the diameter of a circle What is the length of the diameter? What is the center point C of the circle? Given the point C you found in part (b), state the point symmetric to C about the x-axis

May 24, 2018

$d = \sqrt{32} = 4 \sqrt{2} \approx 5.66$
center, $C = \left(- 7 , 4\right)$
symmetric point about $x$-axis: $\left(- 7 , - 4\right)$

#### Explanation:

Given: endpoints of the diameter of a circle: $\left(- 9 , 2\right) , \left(- 5 , 6\right)$

Use the distance formula to find the length of the diameter: $d = \sqrt{{\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({x}_{2} - {x}_{1}\right)}^{2}}$

$d = \sqrt{{\left(- 9 - - 5\right)}^{2} + {\left(2 - 6\right)}^{2}} = \sqrt{16 + 16} = \sqrt{32} = \sqrt{16} \sqrt{2} = 4 \sqrt{2} \approx 5.66$

Use the midpoint formula to find the center: $\left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{1}}{2}\right)$:

$C = \left(\frac{- 9 + - 5}{2} , \frac{2 + 6}{2}\right) = \left(- \frac{14}{2} , \frac{8}{2}\right) = \left(- 7 , 4\right)$

Use the coordinate rule for reflection about the $x$-axis $\left(x , y\right) \to \left(x , - y\right)$:

$\left(- 7 , 4\right)$ symmetric point about $x$-axis: $\left(- 7 , - 4\right)$

May 24, 2018

1) $4 \sqrt{2}$ units.
2) $\left(- 7 , 4\right)$
3) $\left(7 , 4\right)$

#### Explanation:

Let the point A be $\left(- 9 , 2\right)$ & Let the point B be $\left(- 5 , 6\right)$

As the points $A$ and $B$ be the endpoints of the diameter of the circle. Hence, the distance $A B$ be length of the diameter.

Length of the diameter$= \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Length of the diameter$= \sqrt{{\left(- 5 + 9\right)}^{2} + {\left(6 - 2\right)}^{2}}$

Length of the diameter$= \sqrt{{\left(4\right)}^{2} + {\left(4\right)}^{2}}$

Length of the diameter$= \sqrt{32}$

Length of the diameter$= 4 \sqrt{2}$ units.

The centre of the circle is the midpoints of the endpoints of the diameter.

So, by midpoints formula,

${x}_{0} = \frac{{x}_{1} + {x}_{2}}{2}$ & ${y}_{0} = \frac{{y}_{1} + {y}_{2}}{2}$

${x}_{0} = \frac{- 9 - 5}{2}$ & ${y}_{0} = \frac{2 + 6}{2}$

${x}_{0} = \frac{- 14}{2}$ & ${y}_{0} = \frac{8}{2}$

${x}_{0} = - 7$ & ${y}_{0} = 4$

Co-ordinates of the centre$\left(C\right)$= $\left(- 7 , 4\right)$

The point symmetric to C about the x-axis has co-ordinates =$\left(7 , 4\right)$