Question

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

what is the equation in graphing form

a) (x-6)^2+(y-6)^2=27
b) (x-6)^2+y^29
c) x^2+(y+6)^2=9
d) x^2+(y-6)^2=9

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