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

Explanation:

Step1: Find value to complete square for y

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

Step2: Add value to both sides of equation

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

Step3: Rewrite $y$ - terms as perfect - square

The expression $y^{2}-4y + 4$ can be written as $(y - 2)^2$. So the equation becomes $(x - 7)^2+(y - 2)^2=6$.

Answer:

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