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

Answer:

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

Step-by-step 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 $(\frac{b}{2})^2$ 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 $(x-4)^2+(y-3)^2=4$.