Question

select the correct answer. which point lies on the circle represented by the equation ( x^2 + (y - 12)^2 = 25^2 )? a. (20, -3) b. (-7, 24) c. (0, 13) d. (-25, -13)

Explanation:

Step1: Recall circle equation

The standard form of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h,k)\) is the center and \(r\) is the radius. Here, the equation is \(x^2+(y - 12)^2 = 25^2\), so center \((0,12)\), radius \(25\).

Step2: Check point A (20, -3)

Substitute \(x = 20\), \(y=-3\) into the equation:
\(20^2+(-3 - 12)^2=400+(-15)^2=400 + 225 = 625\), and \(25^2 = 625\)? No, \(625
eq625\)? Wait, \(25^2=625\), but \(400 + 225 = 625\)? Wait, no, \(25^2 = 625\), but the left side is \(20^2+( - 15)^2=400 + 225 = 625\), but the equation is \(x^2+(y - 12)^2 = 25^2=625\). Wait, but let's check other points.

Step3: Check point B (-7,24)

Substitute \(x=-7\), \(y = 24\):
\((-7)^2+(24 - 12)^2=49+144 = 193
eq625\).

Step4: Check point C (0,13)

Substitute \(x = 0\), \(y = 13\):
\(0^2+(13 - 12)^2=0 + 1 = 1
eq625\).

Step5: Check point D (-25,-13)

Substitute \(x=-25\), \(y=-13\):
\((-25)^2+(-13 - 12)^2=625+(-25)^2=625 + 625 = 1250
eq625\). Wait, maybe I made a mistake with point A. Wait, the equation is \(x^2+(y - 12)^2 = 25^2\). Let's re - check point A: \(x = 20\), \(y=-3\). Then \((y - 12)=-15\), so \(x^2+(y - 12)^2=20^2+(-15)^2=400 + 225 = 625=25^2\). Oh, I miscalculated earlier. Wait, but let's check the distance from the center \((0,12)\) to point A (20, - 3). The distance formula is \(\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). So \(\sqrt{(20 - 0)^2+(-3 - 12)^2}=\sqrt{400 + 225}=\sqrt{625}=25\), which is equal to the radius. Wait, but let's check point B again. For point B (-7,24): distance from (0,12) is \(\sqrt{(-7 - 0)^2+(24 - 12)^2}=\sqrt{49 + 144}=\sqrt{193}\approx13.89
eq25\). Point C (0,13): distance from (0,12) is \(\sqrt{(0 - 0)^2+(13 - 12)^2}=1
eq25\). Point D (-25,-13): distance from (0,12) is \(\sqrt{(-25 - 0)^2+(-13 - 12)^2}=\sqrt{625 + 625}=\sqrt{1250}\approx35.35
eq25\). Wait, but point A: distance is 25, so it lies on the circle. Wait, but maybe I misread the options. Wait, the equation is \(x^2+(y - 12)^2 = 25^2\). Let's re - check point A: \(x = 20\), \(y=-3\). Then \(x^2=400\), \((y - 12)^2=(-15)^2 = 225\), sum is \(625 = 25^2\). So point A lies on the circle. Wait, but maybe the original equation was \(x^2+(y - 12)^2 = 25^2\), so radius 25, center (0,12). So point A (20, - 3) is on the circle.

Wait, but let's check again. Maybe I made a mistake with point A. Wait, the problem's equation: \(x^2+(y - 12)^2 = 25^2\). So for a point \((x,y)\) to be on the circle, \(x^2+(y - 12)^2\) must equal \(25^2 = 625\).

For point A (20, - 3):
\(20^2+( - 3 - 12)^2=400+( - 15)^2=400 + 225 = 625\), which equals \(25^2\). So point A is on the circle.

Wait, but let's check the other points again. Point B (-7,24): \((-7)^2+(24 - 12)^2=49 + 144 = 193
eq625\). Point C (0,13): \(0^2+(13 - 12)^2=1
eq625\). Point D (-25,-13): \((-25)^2+( - 13 - 12)^2=625 + 625 = 1250
eq625\). So the correct point is A (20, - 3).

Answer:

A. (20, - 3)