Question

The point (-4,-3) lies on a circle whose center is at (0,6). How do you find an equation of this circle?

Answer 1

`x^2+(y-6)^2=109`

Explanation:

If the circle has a center at `(0,6)` and `(-4,-3)` is a point on its circumference,
then it has a radius of:
`color(white)("XXX")r=sqrt((0-(-3))^2+(6-(-4))^2)=sqrt(109)`

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

In this case we have
`color(white)("XXX")x^2+(y-6)^2=109`
graph{x^2+(y-6)^2=109 [-14.24, 14.23, -7.12, 7.11]}

Answer 2

`x^2+y^2+8x+6y-72=0`

Explanation:

It means that `(-4,-3)` is center and radius is distance between `(-4,-3)` and `(0,6)`. The radius is hence given by

`sqrt{(0-(-4))^2+(6-(-3))^2)` or `sqrt(16+81)` or `sqrt87`

Hence equation of circle is

`(x-(-4))^2+(y-(-3^2))=87` or

`(x+4)^2+(y+3)^2=87`

`x^2+8x+16+y^2+6y+9=87` or

`x^2+y^2+8x+6y+16+9-87=0` or

`x^2+y^2+8x+6y-72=0`