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

- anonymous

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- 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

Ok so I have to put (1,7) into that equation you gave me??

- 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

Here is another way of solving.

- mathstudent55

Here is a circle with two points on it and a chord drawn through them.
|dw:1435785536910:dw|

- mathstudent55

|dw:1435785591709:dw|

- mathstudent55

The perpendicular bisector of the chord passes through the center of the circle.
|dw:1435785640579:dw|

- mathstudent55

You are given three points on the circle.
|dw:1435785694244:dw|

- mathstudent55

Use a chord again and a perpendicular bisector again:
|dw:1435785724236:dw|

- anonymous

Ok so how can I figure out the equation by using that method?

- mathstudent55

Where the perpendicular bisectors of the chords intersect is the center of the circle.

- mathstudent55

Ok. Let's start.
|dw:1435785943459:dw|

- mathstudent55

|dw:1435786061026:dw|

- mathstudent55

|dw:1435786091850:dw|

- 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

Ok

- mathstudent55

1. Find the slope of the segment
\(m = \dfrac{6 - 7}{8 - 1} = \dfrac{-1}{7} = -\dfrac{1}{7} \)

- mathstudent55

The slope of the perpendicular bisector is m = 7

- anonymous

ok

- mathstudent55

The perpendicular bisector passes through the midpoint of the segment, so we need to find the midpoint of the segment.

- anonymous

Ok

- mathstudent55

midpoint = \((\dfrac{1 + 8}{2}, \dfrac{7 + 6}{2}) = (\dfrac{9}{2}, \dfrac{13}{2})\)
|dw:1435786426317:dw|

- mathstudent55

We have a point and the slope, so we can find the equation of that perpendicular bisector.

- 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

Now we need to choose another set of two points on the circle and do the same.

- anonymous

Ok so how would I plug that in this equation?
x^2 + y^2__x__y=0

- mathstudent55

|dw:1435786849703:dw|

- mathstudent55

|dw:1435786887498:dw|

- mathstudent55

Now we have the chord in black in the last figure above.
We need its slope and its midpoint.

- anonymous

ok

- 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

We need the coordinates of the midpoint of the chord.
midpoint \(= (\dfrac{7 + 1}{2}, \dfrac{-1 + 7}{2}) = (4, 3)\)

- 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

ok

- 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

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

Ok

- mathstudent55

|dw:1435787686770:dw|

- 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

All we need now the to find the radius, so we can find r in the equation of the circle.

- anonymous

Ok

- 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

Ok

- 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

ok

- anonymous

So it would be 5 right?

- 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

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

Ok

- mathstudent55

|dw:1435788411933:dw|

- anonymous

Thank you!

- mathstudent55

You're welcome.

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