Standard Form of Equation of Circle

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 1, Section 1.4

Find the standard form of the equation of each circle.

36. Center (1, 0) and containing the point (-3, 2).

40. With endpoints of a diameter at (4, 3) and (0, 1).

[Math Tutor]

For both of these you just need the center and the radius, then plug into [tex](x - h)^2 + (y - k)^2 = r^2[/tex].

36. The center is [tex](1, 0)[/tex] and the circle passes through [tex](-3, 2)[/tex], so the radius is the distance between those two points:
[tex]r^2 = (-3 - 1)^2 + (2 - 0)^2 = 16 + 4 = 20[/tex]
No need to take the square root, since the formula wants [tex]r^2[/tex] anyway.
Answer: [tex](x - 1)^2 + y^2 = 20[/tex]

40. The endpoints of a diameter are given, so the center is just the midpoint:
[tex]\left(\frac{4 + 0}{2}, \frac{3 + 1}{2}\right) = (2, 2)[/tex]
The radius is the distance from the center to either endpoint:
[tex]r^2 = (4 - 2)^2 + (3 - 2)^2 = 4 + 1 = 5[/tex]
Answer: [tex](x - 2)^2 + (y - 2)^2 = 5[/tex]

Quick sanity check on #40: the full diameter has length [tex]\sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}[/tex], so [tex]r = \sqrt{5}[/tex], which matches.

[nycmath]

Math Tutor,

Your effort and reply is truly appreciated. Amazing details in all your answers.

[zoryn]

Precalculus
Michael Sullivan
Edition 10
Chapter 1, Section 1.4

Find the standard form of the equation of each circle.

36. Center (1, 0) and containing the point (-3, 2).

40. With endpoints of a diameter at (4, 3) and (0, 1).

for the first one just plug the given center and find r² from the point it goes through. second one midpoint of the diameter ends is the center, then r² from there to either end.

[Eigenvalue]

d=[tex]\sqrt{(-3-1)²+(2)²}[/tex]=[tex]\sqrt{20}[/tex]
(x-1)²+y²=20

[nycmath]

For both of these you just need the center and the radius, then plug into [tex](x - h)^2 + (y - k)^2 = r^2[/tex].

36. The center is [tex](1, 0)[/tex] and the circle passes through [tex](-3, 2)[/tex], so the radius is the distance between those two points:
[tex]r^2 = (-3 - 1)^2 + (2 - 0)^2 = 16 + 4 = 20[/tex]
No need to take the square root, since the formula wants [tex]r^2[/tex] anyway.
Answer: [tex](x - 1)^2 + y^2 = 20[/tex]

40. The endpoints of a diameter are given, so the center is just the midpoint:
[tex]\left(\frac{4 + 0}{2}, \frac{3 + 1}{2}\right) = (2, 2)[/tex]
The radius is the distance from the center to either endpoint:
[tex]r^2 = (4 - 2)^2 + (3 - 2)^2 = 4 + 1 = 5[/tex]
Answer: [tex](x - 2)^2 + (y - 2)^2 = 5[/tex]

Quick sanity check on #40: the full diameter has length [tex]\sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}[/tex], so [tex]r = \sqrt{5}[/tex], which matches.

Fabulous math work and discipline.

[nycmath]

d=[tex]\sqrt{(-3-1)²+(2)²}[/tex]=[tex]\sqrt{20}[/tex]
(x-1)²+y²=20

Would it be possible for you to provide an explanation of your steps for learning and clarity?

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 1, Section 1.4

Find the standard form of the equation of each circle.

36. Center (1, 0) and containing the point (-3, 2).

40. With endpoints of a diameter at (4, 3) and (0, 1).

for the first one just plug the given center and find r² from the point it goes through. second one midpoint of the diameter ends is the center, then r² from there to either end.

Thanks but next time please provide a step by step reply showing how to tackle the problem(s).

[Eigenvalue]

d=[tex]\sqrt{(-3-1)²+(2)²}[/tex]=[tex]\sqrt{20}[/tex]
(x-1)²+y²=20

Would it be possible for you to provide an explanation of your steps for learning and clarity?

Yes, the point (-3,2) has a distance of d,or the radius, from the circle; all points on the circle are equidistant from the center of the circle.

[nycmath]

For both of these you just need the center and the radius, then plug into [tex](x - h)^2 + (y - k)^2 = r^2[/tex].

36. The center is [tex](1, 0)[/tex] and the circle passes through [tex](-3, 2)[/tex], so the radius is the distance between those two points:
[tex]r^2 = (-3 - 1)^2 + (2 - 0)^2 = 16 + 4 = 20[/tex]
No need to take the square root, since the formula wants [tex]r^2[/tex] anyway.
Answer: [tex](x - 1)^2 + y^2 = 20[/tex]

40. The endpoints of a diameter are given, so the center is just the midpoint:
[tex]\left(\frac{4 + 0}{2}, \frac{3 + 1}{2}\right) = (2, 2)[/tex]
The radius is the distance from the center to either endpoint:
[tex]r^2 = (4 - 2)^2 + (3 - 2)^2 = 4 + 1 = 5[/tex]
Answer: [tex](x - 2)^2 + (y - 2)^2 = 5[/tex]

Quick sanity check on #40: the full diameter has length [tex]\sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}[/tex], so [tex]r = \sqrt{5}[/tex], which matches.

Another step by step concise reply for my file notes. Thanks.

[nycmath]

d=[tex]\sqrt{(-3-1)²+(2)²}[/tex]=[tex]\sqrt{20}[/tex]
(x-1)²+y²=20

Would it be possible for you to provide an explanation of your steps for learning and clarity?

Yes, the point (-3,2) has a distance of d,or the radius, from the circle; all points on the circle are equidistant from the center of the circle.

I truly would like you to explain each step along the way in your reply moving forward, if you want to. This will help in terms of clarity and provide an opportunity for members to file your math notes for study time.