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}√(14).`

Required:

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

Step-by-step 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}√(14)\text{ in the formula.}`

`(x-0)^2+(y-3)^2=(2√(14))^2`

`(x-0)^2+(y-3)^2=2(√(14))^2`
`Use\text{ \lparen}√(14))^2=14.`
`x^2+(y-3)^2=2^2*14`

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

Final Answer:

The equation in standard form for the circle is,

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