Math Is Fun Forum

mathland

Member

Registered: 2021-03-25

Posts: 444

Standard Form of the Equation

Write the standard form of each equation of the circle with the given
characteristics.

1. Center: (−2, −6); Solution point: (1, −10)

2. Endpoints of a diameter: (11, −5), (3, 15)

Seeking at least the first-two steps for each question.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,640

Re: Standard Form of the Equation

If (a,b) is the centre of the circle and the radius is r then the equation will be

(x-a)^2 + (y-b)^2 = r^2.

This comes from Pythagoras' theorem.

For the first one you have both the centre and a point on the circumference so you can work out r.

For the second the centre will be half way along the diameter and the radius is half the length of the diameter,

Bob

---

Children are not defined by school ...........The Fonz
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Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathland

Member

Registered: 2021-03-25

Posts: 444

Re: Standard Form of the Equation

If (a,b) is the centre of the circle and the radius is r then the equation will be

(x-a)^2 + (y-b)^2 = r^2.

This comes from Pythagoras' theorem.

For the first one you have both the centre and a point on the circumference so you can work out r.

For the second the centre will be half way along the diameter and the radius is half the length of the diameter,

Bob

For the second one, the midpoint formula comes into play, right?

Mathegocart

Member

Registered: 2012-04-29

Posts: 2,226

Re: Standard Form of the Equation

If (a,b) is the centre of the circle and the radius is r then the equation will be

(x-a)^2 + (y-b)^2 = r^2.

This comes from Pythagoras' theorem.

For the first one you have both the centre and a point on the circumference so you can work out r.

For the second the centre will be half way along the diameter and the radius is half the length of the diameter,

Bob

For the second one, the midpoint formula comes into play, right?

Yes, it would.

---

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He will be sorely missed.

mathland

Member

Registered: 2021-03-25

Posts: 444

Re: Standard Form of the Equation

If (a,b) is the centre of the circle and the radius is r then the equation will be

(x-a)^2 + (y-b)^2 = r^2.

This comes from Pythagoras' theorem.

For the first one you have both the centre and a point on the circumference so you can work out r.

For the second the centre will be half way along the diameter and the radius is half the length of the diameter,

Bob

For the second one, the midpoint formula comes into play, right?

Yes, it would.

Ok. Sounds good. I will play with this problem some more when time allows.
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