Question

The center of a circle is located at (3, 8), and the circle has a radius that is 5 units long. What is the general form of the equation for the circle?

A- x2 + y2 − 6x − 16y + 48 = 0
B- x2 + y2 − 6x − 16y − 25 = 0
C- x2 + y2 + 6x + 16y + 48 = 0
D- x2 + y2 + 6x + 16y − 25 = 0

Answer 1

The answer is A. To understand that, you need to put it in order by the x's and the y's to get x^2-6x + y^2-16y = -48. Now complete the square on both the x and the y terms to get (x^2-6x+9) +(y^2-16y+64) = -48+9+64. Rewriting that in vertex form on the left and doing the math on the right gives you
(x-3)^2 + (y-8)^2 = 25, which shows you a center of (3,8) and a radius of 5.

Answer 2

The correct option is A.

Explanation:

The standard from of a circle is

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

Where, (h,k) is center of the circle and r is the radius.

It is given that the center of the circle is (3,8) and the radius of the circle is 5 units.

Substitute h=3, k=8 and r=5 in the above equation .

`(x-3)^2+(y-8)^2=5^2`

`x^2-6x+9+y^2-16y+64=25`
`[\because (x-y)^2=x^2-2xy+y^2]`

`x^2+y^2-6x-16y+9+64=25`

`x^2+y^2-6x-16y+73=25`

Subtract 25 from both the sides.

`x^2+y^2-6x-16y+73-25=0`

`x^2+y^2-6x-16y+48=0`

Therefore the correct option is A.