Question

Which equation represents the circle described? The radius is 2 units. The center is the same as the center of a circle whose equation is x^2+y^2-8x-6y+24=0

Answer 1

x^2 + y^2 - 8x - 6y + 24 = 0
Completing the square:-
(x - 4)^2 - 16 + (y - 3)^2 - 9 + 24 = 0

(x - 4)^2 + (y - 3)^2 = 1

The equation we need has same center (4,3) and radius 2 so its equation is

(x - 4)^2 + (y - 3)^2 = 2^2

Answer is (x - 4)^2 + (y - 3)^2 = 4

Answer 2

The equation of the circle is
`(x-4)^2+(y-3)^2=4`.

Explanation:

The radius of required circle is 2 units. The standard equation of a circle is

`(x-h)^2+(y-k)^2=r^2` .... (1)

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

The given equation of a circle is

`x^2+y^2-8x-6y+24=0`

`(x^2-8x)+(y^2-6y)+24=0`

If we have the expression
`x^2+bx`, then we have to add
`((b)/(2))^2` to make the expression a perfect square.

`(x^2-8x+4^2)+(y^2-6y+3^2)-4^2-3^2+24=0`

`(x-4)^2+(y-3)^2-1=0`

`(x-4)^2+(y-3)^2=1` .... (2)

From (1) and (2), we get

`h=4,k=3,r=1`

It means the center of this circle is (4,3). So, the center of required circle is also (4,3).

The center is (4,3) and radius is 2, therefore the required equation is

`(x-4)^2+(y-3)^2=2^2`

It can also written as

`(x-4)^2+(y-3)^2=4`

Therefore the equation of the circle is
`(x-4)^2+(y-3)^2=4`.