Question

How do you find the equation of a circle with center (7,k), radius 5 and with the point (4,3) on the circle.?

Answer 1

`{:((x-7)^2+(y+1)^2=25),((x-7)^2+(y-7)^2=25):}`

Explanation:

The general form of the equation of a circle with a center at `(h,k)` and radius `r` is

`(x-h)^2+(y-k)^2=r^2`

Since we have a center of `(7,k)` and `r=5`, we get the equation of

`(x-7)^2+(y-k)^2=25`

Now, to determine `k`, we can use the fact that `(4,3)` is a point on the circle:

`(4-7)^2+(3-k)^2=25`

`9+(3-k)^2=25`

`(3-k)^2=16`

Take the square root of both sides. Note that taking the square root will leave a positive and negative version, so we will have two different equations that fit these criteria.

`3-k=+-4`

When `3-k=4`, we see that `k=-1`.

When `3-k=-4`, we see that `k=7`.

Hence the two possible equations for the circle are

`{:((x-7)^2+(y+1)^2=25),((x-7)^2+(y-7)^2=25):}`

graph{((x-7)^2+(y+1)^2-25)((x-7)^2+(y-7)^2-25)((x-4)^2+(y-3)^2-.1)=0 [-18.27, 27.35, -8.16, 14.65]}

The two large circles are the two possible equations. The smaller circle is the point `(3,4)`, which lies on both circles.