The point (6, n) lies on the circle whose equation is (x − 1)2 + (y − 5)2 = 50. Find the values of n

n = 0 or 10

Step-by-step explanation:

From the question,

Equation of the circle = (x-1)²+(y-5)² = 50 Equation 1

Coordinate of the circle = (6,n)

From the question

x = 6, y = n

Substitute the value of x and y into equation 1

(6-1)²+(n-5)² = 50

5²+(n-5)² = 50

25+(n-5)² = 50

(n-5)² = 50-25

(n-5)² = 25.

Solving for n

n-5 = √(25

n-5 = ±5

Either,

n-5 = -5

n = -5+5

n = 0

or

n-5 = +5

n = 5+5

n = 10

n = 3.16

Step-by-step explanation:

Since the point (6, n) lies on the circle then the coordinates of the point make the equation true.

Substitute x = 6 and y = n into the equation

(6 - 1)² + (n - 5)² = 50, that is, expanding using FOIL

5² + n² - 10n + 25= 50 by simplifying

25 + n² - 10n = 50 ( subtracting 50 from both sides and rearrange )

n² - 10n = 0 ← in standard form

n²=10

n = √10

The values of n is 3.16

no not all of them.

step-by-step explanation:

some of them have a square base.