Question

How do you write the equation for a circle with center at (2,-5) and passing through (3,4)?

Answer 1

`(x-2)^2+(y+5)^2=21`

Explanation:

The general equation of a circle is:

`color(blue)(|bar(ul(color(white)(a/a)(x-h)^2+(y-k)^2=r^2color(white)(a/a)|)))`

where:
`x=`x-coordinate
`h=`x-coordinate of centre
`y=`y-coordinate
`k=`y-coordinate of centre
`r=`radius

Start by plugging `(h,k) = (2,-5)` into the equation.

`(x-(2))^2+(y-(-5))^2=r^2`

Replace `(x,y)` with `(3,4)`.

`(3-2)^2+(4-(-5))^2=r^2`

Simplify.

`(1)^2+(9)^2=r^2`

`1+81=r^2`

`21=r^2`

Rewrite the equation by including the centre and the radius.

`(x-2)^2+(y-(-5))^2=21`

`color(green)(|bar(ul(color(white)(a/a)color(black)((x-2)^2+(y+5)^2=21)color(white)(a/a)|)))`