Question

Write the equation of the circle below to complete the square. Also determine the area of the circle.

Answer 1

Answer:
`(x-7)^2+(y+2)^2=11^2` ;
`A=121\pi`

Explanation:

I have never done a problem like this, so this is my first time attempting it.

The standard form for the equation of a circle is:

`(x-a)^2+(y-b)^2=r^2`

Begin by grouping the 'x' terms and 'y' terms:

`(x^2-14x)+(y^2+4y-68)=0`

Complete the square for the second polynomial:

`((b)/(2) )^2=((4)/(2) )^2=4`

`(x^2-14x)+(y^2+4y+4-4-68)=0`

`(x^2-14x)+(y^2+4y+4)-4-68=0`

`(x^2-14x)+(y^2+4y+4)-72=0`

`(x^2-14x)+(y+2)^2-72=0`

Do the exact same process with the other polynomial:

`(x^2-14x-72)+(y+2)^2=0`

Complete the square:

`((14)/(2) )^2=49`

`(x^2-14x+49-49-72)+(y+2)^2=0`

`(x^2-14x+49)-49-72+(y+2)^2=0`

`(x^2-14x+49)-121+(y+2)^2=0`

`(x-7)^2-121+(y+2)^2=0`

Bring the 121 to the other side to get:

`(x-7)^2+(y+2)^2=121`

`(x-7)^2+(y+2)^2=11^2`

The above expression is the equation of our circle in standard form. It is centered at the coordinates (7, -2) and has a radius of 11 units. The area of a circle if given by:

`A=\pi r^2=121\pi`