Question

Determine the equation of the circle with radius 8 and center (−2,−7).

Answer 1

Answer: $\begin{gathered} (x\text{ + 2\rparen}^2\text{ + \lparen y + 7\rparen}^2\text{ = 64} \\ This\text{ is the equation in vertex form} \\ \\ There\text{ is also equation in general form:} \\ x^2\text{ + y}^2+\text{ 4x + 14y - 11 = 0} \end{gathered}$

Explanation:

Given:

radius = 8

Center of circle is (-2, -7)

To find:

The equation of the circle using the given information

To determine the equation of the circle we will apply the formula:

$\begin{gathered} (x\text{ - h\rparen}^2\text{ + \lparen y - k\rparen}^2\text{ = r}^2 \\ Th\text{ equation of circle in vertex form} \end{gathered}$

where center (-2, -7) = (h, k)

h = -2, k = -7, r = 8

substitute the values into the formula:

$(x\text{ - \lparen-2\rparen\rparen}^2\text{ + \lparen y -\lparen- 7\rparen\rparen}^2\text{ = 8}^2$

$\begin{gathered} (x\text{ + 2\rparen}^2\text{ + \lparen y + 7\rparen}^2\text{ = 64} \\ This\text{ is the equation in vertex form} \\ \\ There\text{ is also equation in standard form:} \\ (x+2)(x+2)\text{ + \lparen y+7\rparen\lparen y+7\rparen = 64} \\ x^2\text{ + 4x + 4 + y}^2+\text{ 14y + 49 = 64} \\ x^2\text{ + y}^2\text{ + 4x + 14y + 53 - 64 = 0} \\ x^2\text{ + y}^2+\text{ 4x + 14y - 11 = 0} \end{gathered}$

The question did not specify which of the form. That is why the equation is written in vertex or standard form

$\begin{gathered} (x\text{ + 2\rparen}^2\text{ + \lparen y + 7\rparen}^2\text{ = 64} \\ This\text{ is the equation in vertex form} \\ \\ There\text{ is also equation in general form:} \\ x^2\text{ + y}^2+\text{ 4x + 14y - 11 = 0} \end{gathered}$