Question

How do you find the equation of a circle that passes through (7, -1) and has a center of (-2, 4)?

Answer 1

`(x+2)^2+(y-4)^2=106`

Explanation:

A circle centred on the origin in the `(x,y)` plane has equation `x^2+y^2=R^2`, where `R` is the radius of the circle. To move any curve in `(x,y)` to a different centre `(x_0,y_0)`, simply replace `x` by `x-x_0` and `y` by `y-y_0`.

So here `(x+2)^2+(y-4)^2=R^2`

To find the needed radius, calculate the distance between `(-2,4)` and `(7,-1)`. Distance is `sqrt((x_a-x_b)^2+(y_a-y_b)^2)`, i.e. `sqrt(81+25)=sqrt(106)`, so `R^2=106` and

`(x+2)^2+(y-4)^2=106`
or, multiplying out terms, which may or not be a more useful form to work with
`x^2+4x+y^2-8y=86`

Answer 2

`(x+2)^2+(y-4)^2=106`

Explanation:

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

`color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))`

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

`"the centre "=(-2,4)" and we require the radius"`

`"the radius is the distance from the centre to the point"`
`"on the circle"`

`"calculate r using the "color(blue)"distance formula"`

`•color(white)(x)r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`"let "(x_1,y_1)=(-2,4)" and "(x_2,y_2)=(7,-1)`

`r=sqrt((7+2)^2+(-1-4)^2)=sqrt(81+25)=sqrt106`

`"substitute values into the equation"`

`(x-(-2))^2+(y-4)^2=(sqrt106)^2`

`(x+2)^2+(y-4)^2=106larrcolor(red)"equation of circle"`