Answer:
a) (5, 11)
b) r = 9
c) (-4, 11)
Step-by-step explanation:
The equation of a circle in standard form:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h, k) - center
r - radius
We have the equation:
[tex](x-5)^2+(y-11)^2=81[/tex]
Therefore
a) h = 5, k = 11 → the center (5, 11)
b) r² = 81 → r = √81 = 9 - radius
c) We choose any value of x, but one which h - r ≤ x ≤ h + r
5 - 9 = -4
5 + 9 = 14
-4 ≤ x ≤ 14
Let x = -4. Put to the equation and solve for y:
[tex](-4-5)^2+(y-11)^2=81[/tex]
[tex](-9)^2+(y-11)^2=81[/tex]
[tex]81+(y-11)^2=81[/tex] subtract 81 from both sides
[tex](y-11)^2=0\to y-11=0[/tex] add 11 to both sides
[tex]y=11[/tex]
(-4, 11)
