Question

(x - 16)^2+(y - 6)^2 = 1
translated 4 left, 2 up

Explanation:

Step1: Recall the standard form of a circle equation

The standard form of a circle equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center of the circle and \(r\) is the radius. For the given equation \((x - 16)^2+(y - 6)^2 = 1\), the center is \((16,6)\) and the radius \(r = 1\).

Step2: Apply the translation rules

When a point \((x,y)\) is translated \(a\) units to the left and \(b\) units up, the new coordinates \((x',y')\) are given by \(x'=x - a\) and \(y'=y + b\). Here \(a = 4\) (left - translation) and \(b=2\) (up - translation). For the center \((h,k)=(16,6)\), the new \(h'\) ( \(x\) - coordinate of the center) is \(h'=16-4 = 12\), and the new \(k'\) ( \(y\) - coordinate of the center) is \(k'=6 + 2=8\).

Step3: Write the new circle equation

Since the radius remains the same (\(r = 1\)), substituting \(h'=12\) and \(k'=8\) into the standard form \((x - h')^2+(y - k')^2=r^2\), we get \((x - 12)^2+(y - 8)^2 = 1\).

Answer:

$$(x - 12)^2+(y - 8)^2 = 1$$