Question

What is the equation of the circle with center (−3, 1) that passes through the point (−5, 3)?

Answer 1

Answer:

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

( x , y) - any point on the circle

( a, b) - center of the circle

r^2 - radius of the circle

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

a = -3

b = 1

( -5 , 3) - point - ( x, y)

x = -5

y = 3

Step 1: substitute the center into the equation

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

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

Step 2 : sub the point into the equation

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

( -5 + 3)^2 + ( 3 - 1)^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 + 3)^2 + ( y - 1)^2 = r^2

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

Therefore, the equation of the circle is

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