Question

Which point is on the circle whose equation is x^2 + y^2 = 289

Answer 1

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 $x^{2} +y^{2} = 289.$

Step 2:

When $(x,y) = (-12,12). x^{2} +y^{2} = (-12)^{2} + (12)^{2} = 288.$ 288 ≠ 289.

When $(x,y) = (7,-10). x^{2} +y^{2} = (7)^{2} + (-10)^{2} = 149.$ 149 ≠ 289.

When $(x,y) = (-1,-16). x^{2} +y^{2} = (-1)^{2} + (-16)^{2} = 257.$ 257 ≠ 289.

When $(x,y) = (8,-15). x^{2} +y^{2} = (8)^{2} + (-15)^{2} = 289.$

So the fourth set of values is on the circle.