anonymous
  • anonymous
Write the general equation for the circle that passes through the points: (1, 7) (8, 6) (7, -1) You must include the appropriate sign (+ or -) in your answer. x^2 + y^2__x__y=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathstudent55
  • mathstudent55
The equation of a circle with center (x, y)and radius r is \((x - h)^2 + (y - k)^2 = r^2\) You have three given points. Substitute one point at a time into the equation above to get 3 three equations in three unknowns, h, k, and r. Solve the system of equations and substitute the values of h, k and r in the above equation.
anonymous
  • anonymous
Ok so I have to put (1,7) into that equation you gave me??
mathstudent55
  • mathstudent55
That is one way of doing it but it's a lot of work unless you have an electronic way of solving the system of equations.

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More answers

mathstudent55
  • mathstudent55
Here is another way of solving.
mathstudent55
  • mathstudent55
Here is a circle with two points on it and a chord drawn through them. |dw:1435785536910:dw|
mathstudent55
  • mathstudent55
|dw:1435785591709:dw|
mathstudent55
  • mathstudent55
The perpendicular bisector of the chord passes through the center of the circle. |dw:1435785640579:dw|
mathstudent55
  • mathstudent55
You are given three points on the circle. |dw:1435785694244:dw|
mathstudent55
  • mathstudent55
Use a chord again and a perpendicular bisector again: |dw:1435785724236:dw|
anonymous
  • anonymous
Ok so how can I figure out the equation by using that method?
mathstudent55
  • mathstudent55
Where the perpendicular bisectors of the chords intersect is the center of the circle.
mathstudent55
  • mathstudent55
Ok. Let's start. |dw:1435785943459:dw|
mathstudent55
  • mathstudent55
|dw:1435786061026:dw|
mathstudent55
  • mathstudent55
|dw:1435786091850:dw|
mathstudent55
  • mathstudent55
Let's start with points (1, 7) and (8, 6) We need to find the equation of the perpendicular bisector of the segment with those endpoints.
anonymous
  • anonymous
Ok
mathstudent55
  • mathstudent55
1. Find the slope of the segment \(m = \dfrac{6 - 7}{8 - 1} = \dfrac{-1}{7} = -\dfrac{1}{7} \)
mathstudent55
  • mathstudent55
The slope of the perpendicular bisector is m = 7
anonymous
  • anonymous
ok
mathstudent55
  • mathstudent55
The perpendicular bisector passes through the midpoint of the segment, so we need to find the midpoint of the segment.
anonymous
  • anonymous
Ok
mathstudent55
  • mathstudent55
midpoint = \((\dfrac{1 + 8}{2}, \dfrac{7 + 6}{2}) = (\dfrac{9}{2}, \dfrac{13}{2})\) |dw:1435786426317:dw|
mathstudent55
  • mathstudent55
We have a point and the slope, so we can find the equation of that perpendicular bisector.
mathstudent55
  • mathstudent55
\(y = mx + b\) \(\dfrac{13}{2} = 7(\dfrac{9}{2}) + b\) \(\dfrac{13}{2} = \dfrac{63}{2} + b\) \(b = -25\) \(y = 7x - 25\) This is the equation of the first perpendicular bisector of a chord.
mathstudent55
  • mathstudent55
Now we need to choose another set of two points on the circle and do the same.
anonymous
  • anonymous
Ok so how would I plug that in this equation? x^2 + y^2__x__y=0
mathstudent55
  • mathstudent55
|dw:1435786849703:dw|
mathstudent55
  • mathstudent55
|dw:1435786887498:dw|
mathstudent55
  • mathstudent55
Now we have the chord in black in the last figure above. We need its slope and its midpoint.
anonymous
  • anonymous
ok
mathstudent55
  • mathstudent55
\(slope = m = \dfrac{-1 - 7}{7 - 1} = \dfrac{-8}{6} = -\dfrac{4}{3}\) The slope of the perpendicular bisector is \(\dfrac{3}{4} \)
mathstudent55
  • mathstudent55
We need the coordinates of the midpoint of the chord. midpoint \(= (\dfrac{7 + 1}{2}, \dfrac{-1 + 7}{2}) = (4, 3)\)
mathstudent55
  • mathstudent55
Since we now have the slope of the perpendicular bisector and a point it goes through, we can find the equation of the perpendicular bisector.
anonymous
  • anonymous
ok
mathstudent55
  • mathstudent55
\(y = mx + b\) \(3 = \dfrac{3}{4} \times 4 + b\) \(b = 0\) The eq of the perp bisector is: \(y = \dfrac{3}{4} x \)
mathstudent55
  • mathstudent55
Now we have two equations of the two perpendicular bisectors of chords that we know intersect at the center pf the circle. We solve the system of equations: \(y = 7x - 25\) \(y = \dfrac{3}{4} x\) Since both equations are solved for y, we can equate the right sides and solve for x. \(7x - 25 = \dfrac{3}{4}x \) \(28x - 100 = 3x\) \(25x = 100\) \(x = 4\) \(y = \dfrac{3}{4} x = \dfrac{3}{4} \times 4 = 3\) Now we know the center is (4, 3).
anonymous
  • anonymous
Ok
mathstudent55
  • mathstudent55
|dw:1435787686770:dw|
mathstudent55
  • mathstudent55
We can already fill in a little in the equation of the circle: The center is (h, k), and we know it is (4, 3). \((x - 4)^2 + (y - 3)^2 = r^2\)
mathstudent55
  • mathstudent55
All we need now the to find the radius, so we can find r in the equation of the circle.
anonymous
  • anonymous
Ok
mathstudent55
  • mathstudent55
The radius of a circle is the distance between the center of the circle and any point on the circle. We know the center is (4, 3) Also, we were given three points on the circle. We now need to find the distance between the center and any of the three points we were given to find the radius.
anonymous
  • anonymous
Ok
mathstudent55
  • mathstudent55
Let's use point (8, 6) The distance between point (8, 6) on the circle, and point (4,3) the center is the radius of the circle. We use the distance formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
anonymous
  • anonymous
ok
anonymous
  • anonymous
So it would be 5 right?
mathstudent55
  • mathstudent55
\(d = \sqrt{(8 - 4)^2 + (6 - 3)^2} \) \(d = \sqrt{4^2 + 3^2} \) \(d = \sqrt{16 + 9} \) \(d = \sqrt{25} \) \(d = 5 \)
mathstudent55
  • mathstudent55
Right. The radius is 5. Since the equation of a circle has r^2 on the right side, we can finish now: \((x - 4)^2 + (y - 3)^2 = 5^2\) or if you prefer \((x - 4)^2 + (y - 3)^2 = 25\)
anonymous
  • anonymous
Ok
mathstudent55
  • mathstudent55
|dw:1435788411933:dw|
anonymous
  • anonymous
Thank you!
mathstudent55
  • mathstudent55
You're welcome.

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