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

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Answer 1 · 2018-05-24T04:32:11

$d = sqrt(32) = 4sqrt(2) ~~ 5.66$
center, $C = (-7, 4)$
symmetric point about $x$ -axis: $(-7, -4)$

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

Use the distance formula to find the length of the diameter: $d = sqrt((y_2 - y_1)^2 + (x_2 - x_1)^2)$

$d = sqrt((-9 - -5)^2 + (2 - 6)^2) = sqrt(16 + 16) = sqrt(32) = sqrt(16)sqrt(2) = 4 sqrt(2) ~~ 5.66$

Use the midpoint formula to find the center: $((x_1 + x_2)/2, (y_1 + y_1)/2)$ :

$C = ((-9 + -5)/2, (2 + 6)/2) = (-14/2, 8/2) = (-7, 4)$

Use the coordinate rule for reflection about the $x$ -axis $(x, y) -> (x, -y)$ :

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

Answer 2 · 2018-05-24T04:34:40

1) $4 sqrt(2)$ units.
2) $(-7,4)$
3) $(7,4)$

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

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

Length of the diameter $= sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

Length of the diameter $= sqrt((-5+9)^2+(6-2)^2)$

Length of the diameter $= sqrt((4)^2+(4)^2)$

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,

$x_0= (x_1+x_2)/2$ & $y_0= (y_1+y_2)/2$

$x_0= (-9-5)/2$ & $y_0= (2+6)/2$

$x_0= (-14)/2$ & $y_0= (8)/2$

$x_0= -7$ & $y_0= 4$

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

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