Question

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

Answer 1

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

Step-by-step 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:

$(\frac{b}{2} )^2=(\frac{4}{2} )^2=4$

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

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

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

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

Do the exact same process with the other polynomial:

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

Complete the square:

$(\frac{14}{2} )^2=49$

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

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

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

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

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:

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