Question

Determine the equation of the circle with center (-1,0) containing the point
(-4, √108).

Answer 1

`(x+1)^2+y^2=117`

Explanation:

Equation of the circle: (x - h)² + (y - k)² = r²

(x, y) – any point in the circle

(h, k) – the center of the circle

r – the radius of the circle

(-4, √108) – (x , y)

(-1, 0) – (h, k)

r – ?

`(x-h)^2+(y-k)^2=r^2\\\\((-4)-(-1))^2+((√(108))-(0))^2=r^2\\\\(-4+1)^2+(√(108)-0 )^2=r^2\\\\(-3)^2+(√(108) )^2=r^2\\\\9+108=r^2\\\\r^2=117\\\\r^2=9*13\\\\√(r^2)=√(9*13) \\\\r=3√(13)`

Now that we know what's the r value:

`(x - (-1))^2+(y-0)^2=(3√(13) )^2\\\\(x+1)^2+y^2=9*13\\\\(x+1)^2+y^2=117`

Answer 2

Answer:
`(x+1)^2+y^2=117`

Explanation:

The radius of the circle is
`\sqrt{(-1-(-4))^2 +(√(108)-0)^2}=√(117)`.

Using the center-radius form for the equation of a circle, the equation is
`(x+1)^2+y^2=117`.