Question

1 write an equation of the circle that has the given center and radius.
(1) c(3,0); r = 8

(2) c(0,0); r = 6

2 the center of the circle $x^2 + y^2 - 8x + 14y = 0$ is ______ .
a. (4,7) b. (4,−7) c. (−4,−7) d. (−4,7)

Explanation:

Problem (1)

Step1: Recall circle equation formula

The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h,k)\) is the center and \(r\) is the radius.

Step2: Substitute center and radius

For center \(C(3,0)\) (so \(h = 3\), \(k = 0\)) and \(r = 8\), substitute into the formula: \((x - 3)^2 + (y - 0)^2 = 8^2\).

Step3: Simplify the equation

Simplify to get \((x - 3)^2 + y^2 = 64\).

Step1: Recall circle equation formula

Use the standard form \((x - h)^2 + (y - k)^2 = r^2\) with center \((h,k)\) and radius \(r\).

Step2: Substitute center and radius

For center \(C(0,0)\) (\(h = 0\), \(k = 0\)) and \(r = 6\), substitute: \((x - 0)^2 + (y - 0)^2 = 6^2\).

Step3: Simplify the equation

Simplify to \(x^2 + y^2 = 36\).

Step1: Recall circle equation completion

To find the center, we complete the square for the given circle equation \(x^2 + y^2 - 8x + 14y = 0\).

Step2: Group \(x\) and \(y\) terms

Group as \((x^2 - 8x) + (y^2 + 14y) = 0\).

Step3: Complete the square for \(x\)

For \(x^2 - 8x\), add \((\frac{-8}{2})^2 = 16\). For \(y^2 + 14y\), add \((\frac{14}{2})^2 = 49\). So: \((x^2 - 8x + 16) + (y^2 + 14y + 49) = 16 + 49\).

Step4: Rewrite as standard form

This becomes \((x - 4)^2 + (y + 7)^2 = 65\). The center \((h,k)\) is \((4, -7)\).

Answer:

\((x - 3)^2 + y^2 = 64\)

Problem (2)