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)
Accepted Solution
A:
Answer:Option A) (-14, -4)Step-by-step explanation:we know thatIf a ordered pair lie on the circumference of a circle , then the ordered pair must satisfy the equation of the circlewe have[tex](x+10)^{2}+(y+1)^{2}=25[/tex]Verify each ordered paircase A) we have (-14, -4)substitute the value of x and the value of y in the equation and then compare the results[tex](-14+10)^{2}+(-4+1)^{2}=25[/tex][tex](-4)^{2}+(-3)^{2}=25[/tex][tex]25=25[/tex] ----> is truethereforeThe ordered pair is on the circumference of the circlecase B) we have (4,14)substitute the value of x and the value of y in the equation and then compare the results[tex](4+10)^{2}+(14+1)^{2}=25[/tex][tex](14)^{2}+(15)^{2}=25[/tex][tex]421=25[/tex] ----> is not truethereforeThe ordered pair is not on the circumference of the circlecase C) we have (-14,4)substitute the value of x and the value of y in the equation and then compare the results[tex](-14+10)^{2}+(4+1)^{2}=25[/tex][tex](-4)^{2}+(5)^{2}=25[/tex][tex]41=25[/tex] ----> is not truethereforeThe ordered pair is not on the circumference of the circlecase D) we have (-4,14)substitute the value of x and the value of y in the equation and then compare the results[tex](-4+10)^{2}+(14+1)^{2}=25[/tex][tex](6)^{2}+(15)^{2}=25[/tex][tex]261=25[/tex] ----> is not truethereforeThe ordered pair is not on the circumference of the circle