Equation of a circle whose centre is in $I$ quadrant as $\left(\alpha,\ \beta\right)$ and touches $x-$axis will be:

Given that: A circle centre in I quadrant as $(\alpha,\beta)$ and touches x-axis.

To find: Equation of the circle.

Solution:

Refer image,

$\therefore$ Centre (c)

$=(\alpha,\beta)$

$\because\,CP$ is parallel to y-axis

$\therefore$ Radius (p) $=CP=\sqrt{(x_2-x_2)^2+(y_2-y_1)^2}$

$=\sqrt{(\alpha-\alpha)^2+(0-\beta)^2}$

$=\sqrt{\beta^2}$

$=\beta$

$\therefore$ Equation of the circle is

$(x-\alpha)^2+(y-\beta)^2=\beta^2$

$\Rightarrow x^2-2\alpha x+\alpha^2+y^2-2\beta y+\beta^2=\beta^2$

$\Rightarrow x^2+y^2-2\alpha x-2\beta y+\alpha^2=0$

Hence, the required is

$x^2+y^2-2\alpha x-2\beta y+\alpha^2=0$