Answer:
1. The center is (-5, 6)
The radius is 11
2. The center is (3, -9)
The radius is 6
3. The center of the circle is (-4, -2)
The radius is 3
4. The coordinate of the center of the circle is (2, 2)
The radius of the circle is 4
5. The equation of the circle is (x - 4)² + (y + 3)² = 2²
6. The equation of the circle is (x + 1)² + (x - 10)² = 7²
7. The given point (6, 8) does not lie on the circumference of the circle
8. Yes, the point (√17, 4) lies on the circumference of the circle
9. Please find the attached graph created with the aid of MS Excel
10 Please find the attached graph created with the aid of MS Excel
11. The equation of the circle is (x - a)² + (y - b)² = c²
12. The equation of the circle is x² + y² = 1
Explanation:
1. The equation of the circle is (x + 5)² + (y - 6)² = 121
The standard form of the equation of a circle is (x - h)² + (y - k)² = r²
Where;
The center of the circle = (h, k)
The radius of the circle = r
By comparison we have;
h = -5, k = 6, and r = √121 = 11
The center of the circle, (h, k) = (-5, 6)
The radius of the circle, r = 11
2. (x - 3)² + (x + 9)² = 36
Similar to question 1, above, we have by comparison;
h = 3, k = -9, r = √36 = 6
The center of the circle, (h, k) = (3, -9)
The radius of the circle, r = 6
3. From the graph of the circle on the coordinate plane, we have;
The center of the circle = (-4, -2)
The radius of the circle = -1 - (-4) = 3
4. From the graph of the locus of the circle, we have;
The coordinate of the center of the circle, r = (2, 2)
The length of the radius of the circle = The distance from the center to the point (-2, 2)
Therefore, the radius of the circle, r = 2 - (-2) = 4
5. The given radius of the circle, r = 2
The center of the circle, (h, k) = (4, -3)
The equation of a circle in standard form is (x - h)² + (y - k)² = r²
By comparison, we have the equation of the circle given as follows;
(x - 4)² + (y - (-3))² = (x - 4)² + (y + 3)² = 2²
The equation of the circle is (x - 4)² + (y + 3)² = 2²
6. The given radius of the circle, r = 7
The center of the circle, (h, k) = (-1, 10)
The equation of a circle in standard form is (x - h)² + (y - k)² = r²
By comparison, we have the equation of the circle given as follows;
(x - (-1))² + (y - 10)² = (x + 1)² + (y - 10)² = 7²
The equation of the circle is (x + 1)² + (x - 10)² = 7²
7. The given point = (6, 8)
The radius, r = 4
The center of the circle = (3, 5)
The point lies on the circle if the distance. 'd', from the point to the center of the circle = The radius of the circle
∴ d = √((6 - 3)² + (8 - 5)²) = √((3² + 3²) = √(18) = 3·√2
Given that the distance from the point to the center of the circle, d = 3·√2 ≠ 4 = r, the point (6, 8), does not lie on the circumference of the circle
8. The point = (√17, 4)
Radius, r: 9
The center: (0, -4)
The distance from the point to the center of the circle, d = √((√17 - 0)² + (4 - (-4))²) = 9 = r, the radius of the circle
Therefore, given that the distance of the point from the center of the circle = The radius of the circle, the point, (√17, 4), lies on the circumference of the circle
9. The given equation of the circle is (x - 5)² + (y - 1)² = 49
Comparing with the equation of a circle in standard form, we have;
The center of the circle, (h, k) = (5, 1)
The radius of the circle, r = √49 = 7
Please find the attached graph created with the aid of MS Excel
10. The given equation of the circle is x² + (y - 3)² = 25
The center of the circle, (h, k) = (0, 3)
The radius of the circle, r = √25 = 5
Please find the attached graph created with the aid of MS Excel
11. Given that the center of the circle is (a, b) and the radius of the circle is c, the equation of a circle in standard form, (x - h)² + (y - k)² = r², becomes;
(h, k) = (a, b)
r = c
∴ (x - a)² + (y - b)² = c²
12. The diameter of the circle, d = 2 in.
The center of the circle = The origin with coordinates, (0, 0)
Therefore;
The radius of the circle. r = d/2 = 2 in./2 = 1 in.
The center of the circle, (h, k) = (0, 0)
The equation of the circle is therefore;
(x - 0)² + (y - 0)² = 1²
Which gives the equation of the circle as x² + y² = 1.