Question

A point on the circumference of the circle with the equation of (x+10)²+(y+1)²=25 is?

A) (-14, -4)
B) (4, 14)
C) (-14, 4)
D) (-4, 14)

Answer 1

Option A) (-14, -4)

Explanation:

we know that

If a ordered pair lie on the circumference of a circle , then the ordered pair must satisfy the equation of the circle

we have

`(x+10)^(2)+(y+1)^(2)=25`

Verify each ordered pair

case A) we have (-14, -4)

substitute the value of x and the value of y in the equation and then compare the results

`(-14+10)^(2)+(-4+1)^(2)=25`

`(-4)^(2)+(-3)^(2)=25`

`25=25` ----> is true

therefore

The ordered pair is on the circumference of the circle

case B) we have (4,14)

substitute the value of x and the value of y in the equation and then compare the results

`(4+10)^(2)+(14+1)^(2)=25`

`(14)^(2)+(15)^(2)=25`

`421=25` ----> is not true

therefore

The ordered pair is not on the circumference of the circle

case C) we have (-14,4)

substitute the value of x and the value of y in the equation and then compare the results

`(-14+10)^(2)+(4+1)^(2)=25`

`(-4)^(2)+(5)^(2)=25`

`41=25` ----> is not true

therefore

The ordered pair is not on the circumference of the circle

case D) we have (-4,14)

substitute the value of x and the value of y in the equation and then compare the results

`(-4+10)^(2)+(14+1)^(2)=25`

`(6)^(2)+(15)^(2)=25`

`261=25` ----> is not true

therefore

The ordered pair is not on the circumference of the circle