Question

Write the standard form of the equation and the generalform of the equation of the circle with radius r and center(h,k). Then graph the circle.AY6r=2; (h,k) = (0,2)4The standard form of the equation of this circle is2The general form of the equation of this circle is(Simplify your answer.)-4-2Graph the circle.-2Click toenlargegraph4

Answer 1

$\begin{gathered} 1)standard\colon x^2+(y-2)^2\text{ = 4} \\ 2)\text{ General: }x^2+y^2\text{ - 4y = 0} \end{gathered}$

Explanation:

coordiantes of the circle center = (h, k) = (0, 2)

h = 0, k = 2

r = 2

$\begin{gathered} Standardform\text{ of the equation of circle:} \\ \text{ }(x-h)^2+(y-k)^2=r^2 \end{gathered}$$\begin{gathered} \text{ }(x-0)^2+(y-2)^2=2^2 \\ x^2+(y-2)^2\text{ = 4} \end{gathered}$$\begin{gathered} \text{General form of the equation of circle:} \\ x^2+y^2\text{ }+\text{ 2gx + 2fy + c = 0} \end{gathered}$$\begin{gathered} center\text{ of circle = (-g, -f)} \\ given\text{ }center\text{ of circle = (0, 2)} \\ -g\text{ = 0 } \\ g\text{ = 0} \\ -f\text{ = 2} \\ f\text{ = -2} \end{gathered}$$\begin{gathered} radius\text{ = }\sqrt[]{g^2+f^2-c} \\ 2\text{ = }\sqrt[]{0^2+(-2)^2\text{ + c}} \\ 2\text{ = }\sqrt[]{0+4\text{ + c}} \\ \text{square both sides:} \\ 2^2\text{ = }4\text{ + c} \\ 4\text{ = 4 + c} \\ c\text{ = 4-4 = 0} \end{gathered}$$\begin{gathered} \text{General form of the equation of circle:} \\ x^2+y^2\text{ }+\text{ 2(0)x + 2(-2)y + 0 = 0} \\ x^2+y^2\text{ - 4y = 0} \end{gathered}$

plotting the graph: