Question

Find the equation of a circle with a center at (-1,5) and a point on the circle is (4,4).

Answer 1

`(x+1)^2+(y-5)^2=26`

Explanation:

The standard form of an equation for a circle is:

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

where the center is (h,k) and the radius is r.

We are given (h,k)=(-1,5):

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

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

We can use (x,y)=(4,4) to find the value for
`r^2`.

`(4+1)^2+(4-5)^2=r^2`

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

`25+1=r^2`

`26=r^2`

So the equation for the circle is:

`(x+1)^2+(y-5)^2=26`

Answer 2

(x + 1)^2 + (y - 5)^2 = 26.

Explanation:

The general form of the equation of a circle is:

(x - a)^2 + (y - b)^2 = r^2 where the center is at (a, b) and r is the radius.

Here a = -1 and b = 5.

The radius of this circle is the length from (-1, 5) to (4, 4) so we can work out r^2 which is:

r^2 = (-1-4)^2 + (5-4)^2

r^2 = 26.

So substituting for a, b and r in the general form the equation of the circle is

(x + 1)^2 + (y - 5(^2 = 26.