Question

How do find the standard equation of the circle with center (3,5) and passes through (-1,-2)?

Answer 1

The standard Cartesian form for the equation of a circle is:

`(x-h)^2+(y-k)^2=r^2" [1]"`

where `(x,y)` is any point on the circle, `(h,k)` is the center, and r is the radius.

Explanation:

Substitute the center `(3,5)` into equation [1]:

`(x-3)^2+(y-5)^2=r^2" [2]"`

Substitute the point `(-1,-2)` into equation [2] and solve for r:

`(-1-3)^2+(-2-5)^2=r^2`

`(-4)^2+(-7)^2=r^2`

`16+49=r^2`

`r^2=65`

`r = sqrt65`

Substitute `sqrt65` for r into equation [2]:

`(x-3)^2+(y-5)^2=(sqrt65)^2" [3]"`

Equation [3] is the answer.

Note: Please understand that I write the right side as `(sqrt65)^2`, instead of 65, because anyone asked for the radius will not make the mistake of reporting 65.