Question

1. for the following general equations of circles, complete the square and write the equations of the circle in standard form. then state the center and the radius of the circle.

a. $x^{2}+y^{2}+14x + 12y+36 = 0$
b. $x^{2}+y^{2}-32x - 14y+296 = 0$

Explanation:

Step1: Complete the square for x and y in part a

For the equation $x^{2}+y^{2}+14x + 12y+36 = 0$.
For the x - terms: $x^{2}+14x=(x + 7)^{2}-49$.
For the y - terms: $y^{2}+12y=(y + 6)^{2}-36$.
The equation becomes $(x + 7)^{2}-49+(y + 6)^{2}-36 + 36=0$, which simplifies to $(x + 7)^{2}+(y + 6)^{2}=49$.

Step2: Identify the center and radius in part a

The standard - form of a circle is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center and r is the radius.
For $(x + 7)^{2}+(y + 6)^{2}=49$, the center is $(-7,-6)$ and the radius $r = 7$.

Step3: Complete the square for x and y in part b

For the equation $x^{2}+y^{2}-32x-14y + 296=0$.
For the x - terms: $x^{2}-32x=(x - 16)^{2}-256$.
For the y - terms: $y^{2}-14y=(y - 7)^{2}-49$.
The equation becomes $(x - 16)^{2}-256+(y - 7)^{2}-49+296=0$, which simplifies to $(x - 16)^{2}+(y - 7)^{2}=9$.

Step4: Identify the center and radius in part b

For $(x - 16)^{2}+(y - 7)^{2}=9$, the center is $(16,7)$ and the radius $r = 3$.

Answer:

a. Standard form: $(x + 7)^{2}+(y + 6)^{2}=49$, Center: $(-7,-6)$, Radius: $7$
b. Standard form: $(x - 16)^{2}+(y - 7)^{2}=9$, Center: $(16,7)$, Radius: $3$