Question

How do you write the equation for a circle with center at (2,3) that is tangent to the x-axis?

Answer 1

`(x-2)^2+(y-3)^2=9`

Explanation:

Given that the center of circle is `(x_1, y_1)\equiv(2, 3)`.

The above circle is tangent to the x-axis hence its radius

`r=\text{y-coordinate}=3`

hence the equation of the circle is given by following formula

`(x-x_1)^2+(y-y_1)^2=r^2`

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

`(x-2)^2+(y-3)^2=9`

Answer 2

`(x-2)^2+(y-3)^2=9`

Explanation:

`"the equation of a circle in standard form is"`

`•color(white)(x)(x-a)^2+(y-b)^2=r^2`

`"where "(a,b)" are the coordinates of the centre and r"`
`"is the radius"`

`"here "(a,b)=(2,3)`

`"r is the vertical distance from the x-axis to the centre"`

`r="y-coordinate of centre "=3`

`(x-2)^2+(y-3)^2=9larrcolor(red)"equation of circle"`