Question

How do you write an equation of a circle with center (-2,1) and radius with endpoint at (1,0)?

Answer 1

`(x+2)^2+(y-1)^2=10`

Explanation:

The simplest equation for a circle is:
`color(white)("XXX")X^2+Y^2=R^2`
for a circle with center at `(0,0)` and radius `R`

If we want to shift this so the center is at `(-2,1)`
then the `X` values become `x=X-2color(white)("xx")rarrcolor(white)("xx")X=x+2`
and
the `Y` values become `y=Y-1color(white)("xx")rarrcolor(white)("xx")Y=y-1`
So `X^2+Y^2` becomes `(x+2)^2+(y-1)^2`

The radius is une3ffected by the shift and will remain the same length.
The radius is the distance between the center `(-2,1)` and any point on the circumference; in this case we are given the point `(1,0)`
Using the Pythagorean Theorem this gives us a radius squared of
`color(white)("XXX")R^2=((-2)-1)^2+(1-0)^2=10`

Therefore the equation of our (shifted) circle will be
`color(white)("XXX")(x+2)^2+(y-1)^2=10`

Answer 2

`(x+2)^2+(y-1)^2=10`

Explanation:

`"the standard form of the equation of a circle 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 distance from the centre to the endpoint gives r"`

`"to calculate r use the "color(blue)"distance formula"`

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

`rArrr=sqrt((-2-1)^2+(1-0)^2)=sqrt10`

`(x-(-2))^2+(y-1)^2=(sqrt10)^2`

`rArr(x+2)^2+(y-1)^2=10larrcolor(red)"equation of circle"`