Question

raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the circle. Write the equation of the circle.

Answer 1

Equation:

`{x}^(2) + {y}^(2) + 2x - 2y - 35= 0`

The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.

Explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

The center of this circle is the midpoint of (-3, 4) and (5, -2).

We use the midpoint formula:

`( (x_1+x_2)/(2), (y_1+y_2,)/(2) )`

Plug in the points to get:

`( ( - 3+5)/(2), ( - 2+4)/(2) )`

`( ( -2)/(2), ( 2)/(2) )`

`( - 1, 1)`

We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:

`r = \sqrt{ {(x_2-x_1)}^(2) + {(y_2-y_1)}^(2) }`

`r = \sqrt{ {(5 - - 1)}^(2) + {( - 2- - 1)}^(2) }`

`r = \sqrt{ {(6)}^(2) + {( - 1)}^(2) }`

`r = √( 36+ 1 ) = √(37)`

The equation of the circle is given by:

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

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

`(x- - 1)^2 + (y-1)^2 = {( √(37)) }^(2)`

`(x + 1)^2 + (y - 1)^2 = 37`

We expand to get:

`{x}^(2) + 2x + 1 + {y}^(2) - 2y + 1 = 37`

`{x}^(2) + {y}^(2) + 2x - 2y +2 - 37= 0`

`{x}^(2) + {y}^(2) + 2x - 2y - 35= 0`

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

`{x}^(2) + {0}^(2) + 2x - 2(0) - 35= 0`

`{x}^(2) +2x - 35= 0`

`(x - 5)(x + 7) = 0`

`x = 5 \: or \: x = - 7`

The point (5,0) and (-7,0) lies on the circle.

When x=0

`{0}^(2) + {y}^(2) + 2(0) - 2y - 35= 0`

`{y}^(2) - 2y - 35= 0`

`(y - 7)(y + 5) = 0`

`y = 7 \: or \: y = - 5`

The point (0,-5) and (0,7) lie on this circle.