Question

Find an equation of the circle that has center (-5, 0) and passes through (4, -6).

PLEASE HELP !!!!

Answer 1

Standard form = (x + 5)² + y² = 117

General Form = x² + 10x + y² - 92 = 0

Step-by-step explanation:

(x - h)² + (y - k)² = r²

• center: (h, k)

Find radius using:

`\sf Distance \ between \ two \ points = √((x_2 - x_1)^2 + (y_2 - y_1)^2)`

`\rightarrow \sf radius : √((-5-4)^+(0-(-6))^2) \ = \ 3√(13) \ \ units`

Find equation inserting values: Given center: (-5, 0)

⇒ (x - (-5))² + (y - 0)² = (3√13)²

⇒ (x + 5)² + y² = 117

⇒ x² + 10x + 25 + y² - 117 = 0

⇒ x² + 10x + y² - 92 = 0

Answer 2

`(x+5)^2+y^2=117`

Step-by-step explanation:

Equation of a circle

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

(where (a, b) is the center and r is the radius)

Given:

• center = (-5, 0)

`\implies (x-(-5))^2+(y-0)^2=r^2`

`\implies (x+5)^2+y^2=r^2`

To find r², input the coordinates of the given point (4, -6) into the equation:

`\implies (4+5)^2+(-6)^2=r^2`

`\implies 81+36=r^2`

`\implies r^2=117`

Therefore, the equation of the circle is:

`\implies (x+5)^2+y^2=117`