Respuesta :
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
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:
The equation of the circle is [tex](x-4)^2+(y-3)^2=4[/tex].
Step-by-step explanation:
The radius of required circle is 2 units. The standard equation of a circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex] .... (1)
where (h,k) is center of the circle and r is radius.
The given equation of a circle is
[tex]x^2+y^2-8x-6y+24=0[/tex]
[tex](x^2-8x)+(y^2-6y)+24=0[/tex]
If we have the expression [tex]x^2+bx[/tex], then we have to add [tex](\frac{b}{2})^2[/tex] to make the expression a perfect square.
[tex](x^2-8x+4^2)+(y^2-6y+3^2)-4^2-3^2+24=0[/tex]
[tex](x-4)^2+(y-3)^2-1=0[/tex]
[tex](x-4)^2+(y-3)^2=1[/tex] .... (2)
From (1) and (2), we get
[tex]h=4,k=3,r=1[/tex]
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
[tex](x-4)^2+(y-3)^2=2^2[/tex]
It can also written as
[tex](x-4)^2+(y-3)^2=4[/tex]
Therefore the equation of the circle is [tex](x-4)^2+(y-3)^2=4[/tex].
