Answer to Question #307433 in Calculus for Sands

Question

1. Determine the center and radius given the equation of a circle in standard form.

1. (x-5)+(y + 4)² = 64

2. (x + 9) +(y-7)² = 121

3. x² + (y+6)² = 4

4. (x-1)² + y² = 1

• II. Convert the equation of a circle in standard form to general form or vice versa.

1. (x-2)² + (y+6)² = 4 2. (x+6)² + (y + 4)² = 36

3. x² + y² + 4x-2y-4=0

Answer 1

Solution (1)

$\begin{array}{l} \left( {x - 5} \right) + \left( {y{\rm{ }} + {\rm{ }}4} \right){\rm{ }} = {\rm{ }}64\\ {\left( {x - 5} \right)^2} + {\left( {y - \left( { - 4} \right)} \right)^2} = {8^2} \end{array}\$ Center is $(5,-4)$ and Radius $8$

$\begin{array}{l} {\left( {x + 9} \right)^2}{\rm{ }} + \left( {y - 7} \right){\rm{ }} = {\rm{ }}121\\ {\left( {x - \left( { - 9} \right)} \right)^2}{\rm{ }} + \left( {y - 7} \right){\rm{ }} = {11^2} \end{array}\$ Center is $(-9,7)$ and Radius $11$

$\begin{array}{l} {\left( {x - 0} \right)^2} + {\left( {y + 6} \right)^2} = 4\\ {\left( {x - 0} \right)^2} + {\left( {y - \left( { - 6} \right)} \right)^2} = {2^2} \end{array}\$ Center is $(0,-6)$ and Radius $2$

$\begin{array}{l} {\left( {x - 1} \right)^2} + {\rm{ }}{y^2}{\rm{ }} = 1\\ {\left( {x - 1} \right)^2} + {\left( {y - 0} \right)^2} = {1^2} \end{array}\$ Center is $(1, 0)$ and Radius $1$

Solution (2)

$\begin{array}{l} {\left( {x - 2} \right)^2} + {\rm{ }}{\left( {y + 6} \right)^2} = {\rm{ }}4\\ {x^2} - 4x + 4 + {y^2} + 12y + 36 = 4\\ {x^2} + {y^2} - 4x + 12y + 28 = 0 \end{array}\$ The general form is ${x^2} + {y^2} - 4x + 12y + 28 = 0$

$\begin{array}{l} {\left( {x + 6} \right)^2} + {\left( {y + 4} \right)^2} = 36\\ {x^2} + 12x + 36 + {y^2} + 8y + 16 = 36\\ {x^2} + {y^2} + 12x + 8y + 16 = 0 \end{array}\$ The general form is ${x^2} + {y^2} + 12x + 8y + 16 = 0$

${x^2} + {y^2} + 4x - 2y - 4 = 0$ This is already in the general form