Question

find two functions defined implicitly by this equation.
$(x - 4)^2 + (y + 3)^2 = 1$
$y = pmsqrt{? - (x - space)^2 + space}$

Explanation:

Step1: Isolate the \((y + 3)^2\) term

We start with the given equation \((x - 4)^2+(y + 3)^2 = 1\). To isolate the term with \(y\), we subtract \((x - 4)^2\) from both sides of the equation.
\[
(y + 3)^2=1-(x - 4)^2
\]

Step2: Take the square root of both sides

Taking the square root of both sides to solve for \(y + 3\), we get:
\[
y + 3=\pm\sqrt{1-(x - 4)^2}
\]

Step3: Solve for \(y\)

Now, we subtract 3 from both sides of the equation to solve for \(y\):
\[
y=\pm\sqrt{1-(x - 4)^2}- 3
\]
Which can be rewritten as \(y=\pm\sqrt{1-(x - 4)^2}+(- 3)\)

Answer:

The values in the boxes are \(1\), \(4\), and \(-3\) respectively. So the equation is \(y=\pm\sqrt{\boldsymbol{1}}-(x - \boldsymbol{4})^2+\boldsymbol{(-3)}\) (or in the form given in the problem \(y=\pm\sqrt{\boldsymbol{1}}-(x - \boldsymbol{4})^2+\boldsymbol{(-3)}\) which is equivalent to \(y=\pm\sqrt{1-(x - 4)^2}-3\))