Question

E center and radius of the circle represented by the equati x^(2)+y^(2)-8x-12y+27=0

Answer 1

Answer: E(4,6), R=5.

Explanation:

Circle equation

`\boxed{(x-a)^2+(y-b)^2=R^2}`,

circle center - (a,b), circle radius - R.

Hence,

`x^2+y^2-8x-12y+27=0\\\\x^2-8y+y^2-12y+27=0\\\\x^2-2*x*4+4^2+y^2-2*x*6+6^2+27-4^2-6^2=0\\\\(x^2-8x+16)+(y^2-12x+36)+27-16-36=0\\\\(x-4)^2+(y-6)^2-25=0\\\\(x-4)^2+(y-6)^2=25\\\\(x-4)^2+(y-6)^2=5^2\\\\Hence,\ a=4,\ b=6,\ R=5\\\\So,\ E(4,6),\ R=5`