Question

1. center: (0, 0); radius: 7
2. center: (-4, 5); radius: 2
3. center: (1, -3); radius: √11
4. center: (3, 8); solution point: (-9, 13)

(1, -10)

Explanation:

Step1: Recall circle equation formula

The standard equation of a circle with center $(h,k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.

Step2: Solve Exercise 58

Substitute $(h,k)=(0,0)$, $r=7$:
$(x-0)^2 + (y-0)^2 = 7^2$
Simplify: $x^2 + y^2 = 49$

Step3: Solve Exercise 59

Substitute $(h,k)=(-4,5)$, $r=2$:
$(x-(-4))^2 + (y-5)^2 = 2^2$
Simplify: $(x+4)^2 + (y-5)^2 = 4$

Step4: Solve Exercise 60

Substitute $(h,k)=(1,-3)$, $r=\sqrt{11}$:
$(x-1)^2 + (y-(-3))^2 = (\sqrt{11})^2$
Simplify: $(x-1)^2 + (y+3)^2 = 11$

Step5: Solve Exercise 61

First calculate radius using distance formula:
$r = \sqrt{(-9-3)^2 + (13-8)^2} = \sqrt{(-12)^2 + 5^2} = \sqrt{144+25} = \sqrt{169} = 13$
Substitute $(h,k)=(3,8)$, $r=13$:
$(x-3)^2 + (y-8)^2 = 13^2$
Simplify: $(x-3)^2 + (y-8)^2 = 169$

Step6: Solve partial Exercise 62

First calculate radius using distance formula (center $(-2,-6)$, point $(1,-10)$):
$r = \sqrt{(1-(-2))^2 + (-10-(-6))^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5$
Substitute $(h,k)=(-2,-6)$, $r=5$:
$(x-(-2))^2 + (y-(-6))^2 = 5^2$
Simplify: $(x+2)^2 + (y+6)^2 = 25$

Answer:

1. $x^2 + y^2 = 49$
2. $(x+4)^2 + (y-5)^2 = 4$
3. $(x-1)^2 + (y+3)^2 = 11$
4. $(x-3)^2 + (y-8)^2 = 169$
5. $(x+2)^2 + (y+6)^2 = 25$