Q.37 - Precalculus Enhanced with Graphing Utilities

Question

Find the standard form of the equation of each circle.

Center (2, 3) and tangent to the x-axis

Step-by-Step Solution

Answer

The standard form of the equation of circle at the center(2,3) and tangent to the x-axis is ${(x-2)}^2+{(y-3)}^2=9$ and the generalized form of the equation of the circle is $x^2+y^2-4x-6y+4=0$

Step 1. Given information

Given that circle at the center (2,3)  and tangent to x-axis.

 we have to find the standard form of the equation for the circle

Step 2. Find radius r of the circle

To formulate the equation of the circle, first, we need to find out the radius. Here circle centered at (2,3 ) and tangent to x-axis, tangent to x-axis means touching the x-axis, so the distance from the x-axis is radius and distance from center is equal is height.so radius is 3.

Step 3. Formation of standard of the equation of the circle

The standard form of the equation of a circle is of radius r with center at the

origin (h, k)  is ${(x-h)}^2+{(y-k)}^2=r^2$.

Hence  the standard form of the equation of the circle is $$\begin{gathered} {(x-2)}^2+{(y-3)}^2=9 \\ To get generalized form of equation open the paranthesis and solve ,then we get \\ {x}^2+-4x+4+y^2-6x+9-9=0 \\ {x}^2+y^2-4x-6y+4=0 \end{gathered}$$