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lys-0071 [83]
4 years ago
6

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

Mathematics
1 answer:
goldenfox [79]4 years ago
3 0

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.

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