Question

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

Answer 1

Answer:

Option A) (-14, -4)

Step-by-step 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