Question

How do you write the standard from of the equation of the circle given center C ( 6, - 2 ), radius 2?

Answer 1

`(x-6)^2+(y+2)^2=4`

Explanation:

`"the equation of a circle in standard form is"`

`•color(white)(x)(x-a)^2+(y-b)^2=r^2`

`"where "(a,b)" are the coordinates of the centre and r is"`
`"the radius"`

`"here "(a,b)=(6,-2)" and "r=2`

`(x-6)^2+(y+2)^2=4larrcolor(blue)"equation of circle"`

Answer 2

`(x-6)^2+(y+2)^2=4`

Explanation:

Recall that the equation of a circle is given by

`bar( ul(|color(white)(2/2)(x-h)^2+(y-k)^2=r^2color(white)(2/2)|))`, with center `(h,k)` and radius `r`

We know that we are centered at `(6,-2)` and have a radius of `2`, so `h=6, k=-2` and `r^2=4`.

We can plug these into our equation to get

`(x-6)^2+(y+2)^2=4`

Hope this helps!