Question

which of the following points lies on the circle with equation $(x - 1)^2+(y + 2)^2=25$?
a. $(6,-2)$
b. $(5,2)$
c. $(1,-3)$
d. $(1,7)$
which of the following points lies on the circle with equation $(x - 4)^2+(y + 2)^2=36$?
a. $(4,8)$
b. $(4,-2)$
c. $(4,4)$
d. $(0,2)$

Explanation:

Step1: Check point \(a(6, - 2)\) for \((x - 1)^{2}+(y + 2)^{2}=25\)

Substitute \(x = 6\) and \(y=-2\) into the equation \((x - 1)^{2}+(y + 2)^{2}\).
\((6 - 1)^{2}+(-2 + 2)^{2}=5^{2}+0^{2}=25\)

Step2: Check point \(d(9,2)\) for \((x - 4)^{2}+(y + 2)^{2}=36\)

Substitute \(x = 9\) and \(y = 2\) into the equation \((x - 4)^{2}+(y + 2)^{2}\).
\((9 - 4)^{2}+(2 + 2)^{2}=5^{2}+4^{2}=25 + 16=41
eq36\)

Step3: Check point \(b(4,-2)\) for \((x - 4)^{2}+(y + 2)^{2}=36\)

Substitute \(x = 4\) and \(y=-2\) into the equation \((x - 4)^{2}+(y + 2)^{2}\).
\((4 - 4)^{2}+(-2 + 2)^{2}=0^{2}+0^{2}=0
eq36\)

Step4: Check point \(c(4,4)\) for \((x - 4)^{2}+(y + 2)^{2}=36\)

Substitute \(x = 4\) and \(y = 4\) into the equation \((x - 4)^{2}+(y + 2)^{2}\).
\((4 - 4)^{2}+(4 + 2)^{2}=0^{2}+6^{2}=36\)

Step5: Check point \(b(5,2)\) for \((x - 1)^{2}+(y + 2)^{2}=25\)

Substitute \(x = 5\) and \(y = 2\) into the equation \((x - 1)^{2}+(y + 2)^{2}\).
\((5 - 1)^{2}+(2 + 2)^{2}=4^{2}+4^{2}=16 + 16=32
eq25\)

Step6: Check point \(c(1,-3)\) for \((x - 1)^{2}+(y + 2)^{2}=25\)

Substitute \(x = 1\) and \(y=-3\) into the equation \((x - 1)^{2}+(y + 2)^{2}\).
\((1 - 1)^{2}+(-3 + 2)^{2}=0^{2}+(-1)^{2}=1
eq25\)

Step7: Check point \(d(1,7)\) for \((x - 1)^{2}+(y + 2)^{2}=25\)

Substitute \(x = 1\) and \(y = 7\) into the equation \((x - 1)^{2}+(y + 2)^{2}\).
\((1 - 1)^{2}+(7 + 2)^{2}=0^{2}+9^{2}=81
eq25\)

Answer:

For \((x - 1)^{2}+(y + 2)^{2}=25\), the answer is \(a.(6,-2)\).
For \((x - 4)^{2}+(y + 2)^{2}=36\), the answer is \(c.(4,4)\).