How do you find the equation of a circle in standard form given C(-2,8) and r=4?

Question

How do you find the equation of a circle in standard form given C(-2,8) and r=4?

1 Answer

Related questions

How do you write the general form given a circle that passes through the given points. P(-2, 0),...
What is the equation of the circle with a center at $(7, 1)$ $(7 ,1 )$ and a radius of $2$ $2$ ?
What is the equation of the circle with a center at $(3, 1)$ $(3 ,1 )$ and a radius of $2$ $2$ ?
What is the equation of the circle with a center at $(3, 1)$ $(3 ,1 )$ and a radius of $1$ $1$ ?
What is the equation of the circle with a center at $(3, 5)$ $(3 ,5 )$ and a radius of $1$ $1$ ?
What is the equation of the circle with a center at $(3, 5)$ $(3 ,5 )$ and a radius of $6$ $6$ ?
What is the equation of the circle with a center at $(2, 5)$ $(2 ,5 )$ and a radius of $6$ $6$ ?
What is the equation of the circle with a center at $(2, 2)$ $(2 ,2 )$ and a radius of $4$ $4$ ?
What is the equation of the circle with a center at $(2, 2)$ $(2 ,2 )$ and a radius of $3$ $3$ ?
What is the equation of the circle with a center at $(2, 1)$ $(2 ,1 )$ and a radius of $3$ $3$ ?

Impact of this question

1429 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-03T12:07:41

The equation of circle is ${(x + 2)}^{2} + {(y - 8)}^{2} = 16$ $(x+2)^2 +(y-8)^2=16$

Explanation:

The equation of circle in standard form is $(x - h^{2}) + {(y - k)}^{2} = r^{2}$ $(x-h^2) + (y-k)^2 =r^2$ , where $(h, k)$ $(h,k)$ is co-ordinate of centre and $r$ $r$ is the radius of circle.

So The equation of circle is ${(x - (- 2))}^{2} + {(y - 8)}^{2} = 4^{2} or {(x + 2)}^{2} + {(y - 8)}^{2} = 16$ $(x - (-2))^2 + (y-8)^2 =4^2 or (x+2)^2 +(y-8)^2=16$ [Ans]

Science

Math

Humanities

... and beyond

How do you find the equation of a circle in standard form given C(-2,8) and r=4?

1 Answer

Explanation:

Related questions

Impact of this question