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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Here is another way of solving.
Here is a circle with two points on it and a chord drawn through them. |dw:1435785536910:dw|
|dw:1435785591709:dw|
The perpendicular bisector of the chord passes through the center of the circle. |dw:1435785640579:dw|
You are given three points on the circle. |dw:1435785694244:dw|
Use a chord again and a perpendicular bisector again: |dw:1435785724236:dw|
Ok so how can I figure out the equation by using that method?
Where the perpendicular bisectors of the chords intersect is the center of the circle.
Ok. Let's start. |dw:1435785943459:dw|
|dw:1435786061026:dw|
|dw:1435786091850:dw|
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.
Ok
1. Find the slope of the segment \(m = \dfrac{6 - 7}{8 - 1} = \dfrac{-1}{7} = -\dfrac{1}{7} \)
The slope of the perpendicular bisector is m = 7
ok
The perpendicular bisector passes through the midpoint of the segment, so we need to find the midpoint of the segment.
Ok
midpoint = \((\dfrac{1 + 8}{2}, \dfrac{7 + 6}{2}) = (\dfrac{9}{2}, \dfrac{13}{2})\) |dw:1435786426317:dw|
We have a point and the slope, so we can find the equation of that perpendicular bisector.
\(y = mx + b\) \(\dfrac{13}{2} = 7(\dfrac{9}{2}) + b\) \(\dfrac{13}{2} = \dfrac{63}{2} + b\) \(b = -25\) \(y = 7x - 25\) This is the equation of the first perpendicular bisector of a chord.
Now we need to choose another set of two points on the circle and do the same.
Ok so how would I plug that in this equation? x^2 + y^2__x__y=0
|dw:1435786849703:dw|
|dw:1435786887498:dw|
Now we have the chord in black in the last figure above. We need its slope and its midpoint.
ok
\(slope = m = \dfrac{-1 - 7}{7 - 1} = \dfrac{-8}{6} = -\dfrac{4}{3}\) The slope of the perpendicular bisector is \(\dfrac{3}{4} \)
We need the coordinates of the midpoint of the chord. midpoint \(= (\dfrac{7 + 1}{2}, \dfrac{-1 + 7}{2}) = (4, 3)\)
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.
ok
\(y = mx + b\) \(3 = \dfrac{3}{4} \times 4 + b\) \(b = 0\) The eq of the perp bisector is: \(y = \dfrac{3}{4} x \)
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).
Ok
|dw:1435787686770:dw|
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). \((x - 4)^2 + (y - 3)^2 = r^2\)
All we need now the to find the radius, so we can find r in the equation of the circle.
Ok
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.
Ok
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: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
ok
So it would be 5 right?
\(d = \sqrt{(8 - 4)^2 + (6 - 3)^2} \) \(d = \sqrt{4^2 + 3^2} \) \(d = \sqrt{16 + 9} \) \(d = \sqrt{25} \) \(d = 5 \)
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 \((x - 4)^2 + (y - 3)^2 = 25\)
Ok
|dw:1435788411933:dw|
Thank you!
You're welcome.

Not the answer you are looking for?

Search for more explanations.

Ask your own question