Question

Write the standard form of the equation of the circle with center (7,1) that also contains the point (−1,−5).

Enter the equation in simplest terms.
pls no links. my computer can't open

Answer 1

`(x-7)^2+(y-1)^2=100`

Explanation:

1) Find the radius

We can do this by using the distance equation with the centre (7,1) and the given point (-1,-5):

`d=√((x_2-x_1)^2+(y_2-y_1)^2)` where the two points are
`(x_1,y_1)` and
`(x_2,y_2)`

Plug in the points (7,1) and (-1,-5)

`d=√((-1-7)^2+(-5-1)^2)\\d=√((-8)^2+(-6)^2)\\d=√(64+36)\\d=√(100)\\d=10`

Therefore, the radius of the circle is 10 units.

2) Plug the data into the equation of a circle

Equation of a circle (when not centred at the origin):

`(x-h)^2+(y-k)^2=r^2` where the centre is
`(h,k)` and r is the radius

Plug in the centre (7,1) as (h,k)

`(x-7)^2+(y-1)^2=r^2`

Plug in the radius 10

`(x-7)^2+(y-1)^2=10^2\\(x-7)^2+(y-1)^2=100`

I hope this helps!