Question

The point (2,n) lies on the circle whose equation is (x-4)^2+(y-2)^2=40. Find the values of n.

The values of n are ___ or ____

Answer 1

`n = - 4 \: or \: n = 8`

Explanation:

If the point (2,n) lies on the circle whose equation is:

`( {x - 4)}^(2) + {(y - 2)}^(2) = 40`

Then this point must satisfy the equation of the circle:

We substitute x=2 and y=n into the equation to get:

`( {2 - 4)}^(2) + {(n - 2)}^(2) = 40`

We simplify:

`4+ {(n - 2)}^(2) = 40`

This implies that,

`{(n - 2)}^(2) = 40 - 4`

`{(n - 2)}^(2) =36`

Take square root;

`n - 2 = \pm √(36)`

Evaluate:

`n - 2 = \pm6`

`n =2 \pm6`

`n =2 - 6 \: or \: n = 2 + 6`

`n = - 4 \: or \: n = 8`