Question

How do you write the equation of the circle whose centre is at (-5, 3) and which passes through the point (-4, -5)?

Answer 1

`(x+5)^2+(y-3)^2=65`

Explanation:

The general equation for a circle with center `(a,b)` and radius `r` is
`color(white)("XXX")(x-a)^2+(y-b)^2=r^2`

We are given
`color(white)("XXX")a=-5` and
`color(white)("XXX")b=+3`

so we only need to find the radius.
The radius is the distance from the center to any point on the circumference.
Given the center, `(-5,3)`, and a point on the circumference, `(-4,-5)` we can evaluate the radius using the Pythagorean Theorem
`color(white)("XXX")r^2=(-5-(-4))^2+(3-(-5))^2`

`color(white)("XXX=")=(-1)^2+8^2`

`color(white)("XXX=")=65`

Therefore the equation of the circle is
`color(white)("XXX")(x-(-5))^2+(y-3)^2=65`
or, simplifying the first term
`color(white)("XXX")(x+5)^2+(y-3)^2=65`