Question

8. (x - 5) + (y - 11)^2= 81a) find its centerb) its radiusc) one point on the circle​

Answer 1

Answer:

a) (5, 11)

b) r = 9

c) (-4, 11)

Step-by-step explanation:

The equation of a circle in standard form:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have the equation:

$(x-5)^2+(y-11)^2=81$

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:

$(-4-5)^2+(y-11)^2=81$

$(-9)^2+(y-11)^2=81$

$81+(y-11)^2=81$           subtract 81 from both sides

$(y-11)^2=0\to y-11=0$         add 11 to both sides

$y=11$

(-4, 11)