Answer: [tex](x-7)^2+(y+2)^2=11^2[/tex] ; [tex]A=121\pi[/tex]
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:
[tex](x-a)^2+(y-b)^2=r^2[/tex]
Begin by grouping the 'x' terms and 'y' terms:
[tex](x^2-14x)+(y^2+4y-68)=0[/tex]
Complete the square for the second polynomial:
[tex](\frac{b}{2} )^2=(\frac{4}{2} )^2=4[/tex]
[tex](x^2-14x)+(y^2+4y+4-4-68)=0[/tex]
[tex](x^2-14x)+(y^2+4y+4)-4-68=0[/tex]
[tex](x^2-14x)+(y^2+4y+4)-72=0[/tex]
[tex](x^2-14x)+(y+2)^2-72=0[/tex]
Do the exact same process with the other polynomial:
[tex](x^2-14x-72)+(y+2)^2=0[/tex]
Complete the square:
[tex](\frac{14}{2} )^2=49[/tex]
[tex](x^2-14x+49-49-72)+(y+2)^2=0[/tex]
[tex](x^2-14x+49)-49-72+(y+2)^2=0[/tex]
[tex](x^2-14x+49)-121+(y+2)^2=0[/tex]
[tex](x-7)^2-121+(y+2)^2=0[/tex]
Bring the 121 to the other side to get:
[tex](x-7)^2+(y+2)^2=121[/tex]
[tex](x-7)^2+(y+2)^2=11^2[/tex]
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:
[tex]A=\pi r^2=121\pi[/tex]
