Question

What is the equation of the circle with center (1, -1) that passes through the point (5, 7)?

Answer 1

Answer:

Equation of the circle =

80 = ( x - 1)^2 + (y + 1)^2

Step-by-step explanation:

Center = (1 , -1)

Point = ( 5 , 7)

Eqn of a circle is

r^2 = (x - h)^2 + (y - k)^2

We are not given the radius of the circle, fortunately we are provided the information that the circle contains a point ( 5, 7), so we use the above information to find r

Using the center (1, -1)

h - 1

k - -1

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

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

With the point ( 5,7)

x = 5

y = 7

r^2 = ( 5 - 1)^2 + ( 7 + 1)^2

= 4^2 + 8^2

= 16 + 64

= 80

r^2 = 80

r = square root of 80

r = 8.94

The r = 8.94 , which means the equation of the circle is

80 = (x - 1)^2 + ( y + 1)^2