Question

Find an equation of the circle that has center (-1, 3) and passes through (1. 1)

Answer 1

Answer:

(x + 1)² + (y - 3)² = 8

Step-by-step explanation:

radius (r): √(1 - (-1))² + (1 - 3)² = √8 = 2√2

Formula for circle with center (h,k): (x – h)² + (y – k)² = r²

(x + 1)² + (y - 3)² = 8

Answer 2

Answer:

( x + 1)^2 + (y - 3)^2 = 8

Step-by-step explanation:

The equation of a circle with a center and a point

( x - a) ^2 + ( y - b) ^2 = r^2

( a , b) - center of the circle

( x, y) - any point on the circle

r^2 - radius of the circle

We are provided with a center and a point

( -1 , 3) - ( a, b) - center

a = -1

b = 3

( 1 , 1) - ( x, y) - point

x = 1

y = 1

Step1: substitute the center of the circle into the equation

( x - (-1)^2 + ( y - 3)^2 = r^2

(x + 1)^2 + (y - 3)^2 = r^2

Step 2: sub the point into the equation

( x + 1)^2 + (y - 3)^2 = r^2

x - 1

y - 1

( 1 + 1)^2 + ( 1 - 3)^2 = r^2

( 2)^2 + ( -2)^2 = r^2

4 + 4 = r^2

8 = r^2

Step 3 : sub the radius into the equation

( x + 1)^2 + ( y - 3)^2 = r^2

( x + 1)^2 + (y - 3 )^2 = 8

Therefore, the equation of the circle is

( x + 1)^2 + ( y - 3)^2 = 8