Question

Write a equation of a circle with a center of (4,-1) that contains the points (6,0)

Answer 1

To start to solve this problem, we will first plot both points

Since (4,-1) is the center of the circle, a picture of the desired circle would look like this

Note that the radius of the circle corresponds to the distance between this two points. So first, we will find the distance between both points and then we will use the circle's general equation to find the equation of the circle.

Recall that given points (a,b) and (c,d), distance between them is given by the formula

`d=\sqrt[]{(c-a)^2+(d-b)^2}`

In our case, taking a=4, b= -1, c=6 and d=0, we have

`r=d=\sqrt[]{(6-4)^2+(0-(-1))^2}=\sqrt[]{2^2+1^2}=\sqrt[]{5}`

If we raise both sides to the power of 2, we get

`r^2=(\sqrt[]{5})^2=5`

Now, recall that the general equation of a circle centered at point (h,k) of radius r is given by the formula

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

In our case, we know that (4,-1) is the center. So h=4, k= -1 and we calculated that r^2=5. So the equation of the desired circle would be

`(x-4)^2+(y-(-1))^2=5`

or equivalently

`(x-4)^2+(y+1)^2=5`