Answer :
Answer:
The correct statements are B and D.
Step-by-step explanation:
The equation [tex](x-3)^2+(y+1)^2=9[/tex] reminds us of the circle equation [tex]\boxed{(x-a)^2+(y-b)^2=r^2}[/tex], where:
- [tex](a,b)[/tex] = coordinate of the circle's center
- [tex]r[/tex] = radius
In order to find the center and the radius, we convert the equation into the standard form:
[tex](x-3)^2+(y+1)^2=9[/tex]
[tex](x-(3))^2+(y-(-1))^2=3^2[/tex]
Therefore:
- center [tex](a,b)[/tex] = (3, -1)
- radius ([tex]r[/tex]) = 3
Hence, options A and C are incorrect and option B is correct.
for option D:
When the equation intersects the x-axis, the y-value = 0.
[tex](x-3)^2+(y+1)^2=9[/tex] → substitute [tex]y[/tex] with 0
[tex](x-3)^2+(0+1)^2=9[/tex]
[tex](x-3)^2=9-1[/tex]
[tex]x-3=\pm\sqrt{8}[/tex]
[tex]x=\pm2\sqrt{2} +3[/tex]
[tex]x=2\sqrt{2} +3\ or\ -2\sqrt{2} +3[/tex]
Since there are 2 solutions for x-value, then the equations intersects the x-axis twice. (option D is correct)
for option E:
When the equation intersects the y-axis, the x-value = 0.
[tex](x-3)^2+(y+1)^2=9[/tex] → substitute [tex]x[/tex] with 0
[tex](0-3)^2+(y+1)^2=9[/tex]
[tex](y+1)^2=9-9[/tex]
[tex]y+1=0[/tex]
[tex]y=-1[/tex]
Since there is only 1 solution for y-value, then the equations tangent to the y-axis. (option E is incorrect)
