Question

complete the square for y, creating a perfect - square trinomial by adding a value to both sides.
to create a perfect - square trinomial for the y variable, add the square of half the coefficient of y to both sides.
$\left(\frac{4}{2}\
ight)^2=(2)^2=4$✓
add this value to both sides of the equation.
$(x - 7)^2+(y^2 - 4y)=2$
$(x - 7)^2+(y^2 - 4y + 4)=2 + 4$
$(x - 7)^2+(y^2 - 4y + 4)=2$✓✗
the group of terms $(y^2 - 4y + 4)$ is a trinomial that can be written as a perfect square. rewrite the equation.
$(x - 7)^2+(y - 2)^2=6$✓
thus, written in standard form, an equation for the circle is given by the following.
$(x - 7)^2+(y - 2)^2=6$✓

Explanation:

Step1: Identify coefficient of y

The coefficient of $y$ in $y^{2}-4y$ is $- 4$.

Step2: Calculate value to add

Half of the coefficient of $y$ is $\frac{-4}{2}=-2$, and its square is $(-2)^{2}=4$.

Step3: Add value to both sides of equation

Given $(x - 7)^{2}+(y^{2}-4y)=2$, adding 4 to both sides gives $(x - 7)^{2}+(y^{2}-4y + 4)=2 + 4$.

Step4: Rewrite as perfect - square

Since $y^{2}-4y + 4=(y - 2)^{2}$, the equation becomes $(x - 7)^{2}+(y - 2)^{2}=6$.

Answer:

$(x - 7)^{2}+(y - 2)^{2}=6$