Point 4) (8, -15) is on the circle.
Step-by-step explanation:
Step 1:
To determine which point lies on the circle equation, we substitute the different x and y values in the equation and check to see which best satisfies the values in the equation.
The equation of the circle is [tex]x^{2} +y^{2} = 289.[/tex]
Step 2:
When [tex](x,y) = (-12,12). x^{2} +y^{2} = (-12)^{2} + (12)^{2} = 288.[/tex] 288 ≠ 289.
When [tex](x,y) = (7,-10). x^{2} +y^{2} = (7)^{2} + (-10)^{2} = 149.[/tex] 149 ≠ 289.
When [tex](x,y) = (-1,-16). x^{2} +y^{2} = (-1)^{2} + (-16)^{2} = 257.[/tex] 257 ≠ 289.
When [tex](x,y) = (8,-15). x^{2} +y^{2} = (8)^{2} + (-15)^{2} = 289.[/tex]
So the fourth set of values is on the circle.
