Question

Write the equation in standard form for the circle with center (0,3) and radius 2/14

Answer 1

Given:

$\text{The center, \lparen h,k\rparen =\lparen0,3\rparen.}$$The\text{ radius, r=2}\sqrt{14}.$

Required:

We need to find the equation in standard form for the circle.

Explanation:

Consider the equation in standard form for the circle.

$(x-h)^2+(y-k)^2=r^2$$\text{ Substitute h =0, k=3 and r =2}\sqrt{14}\text{ in the formula.}$

$(x-0)^2+(y-3)^2=(2\sqrt{14})^2$

$(x-0)^2+(y-3)^2=2(\sqrt{14})^2$$Use\text{ \lparen}\sqrt{14})^2=14.$$x^2+(y-3)^2=2^2\times14$

$x^2+(y-3)^2=56$

Final Answer:

The equation in standard form for the circle is,

$x^2+(y-3)^2=56.$