Question

The equation of a circle is x^2+y^2-12y+27=0

Answer 1

The equation in graphing form:

$(x-h)^2+(y-k)^2=r^2$

Use: $(*)\ a\pm b)^2=a^2\pm2ab+b^2$

$x^2+y^2-12y+27=0\\\\x^2+y^2-2\cdot y\cdot6+27=0\\\\x^2+\underbrace{y^2-2\cdot y\cdot6+6^2}_{(*)}-6^2+27=0\\\\x^2+(y-6)^2-36+27=0\\\\x^2+(y-6)^2-9=0\ \ \ \ |+9\\\\x^2+(y-6)^2=9$

Answer: d) x^2+(y-6)^2=9