Question

How do you find the general form of circle centred at $(2,3)$ and tangent to x-axis ?

Answer 1

General form for equation of circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$
Here centre is $( - g, - f)$ , radius is $\sqrt {{g^2} + {f^2} - c}$ and $g,f,c$ are three constants.
The equation of circle with centre $(h,k)$ and radius $r$ is
${(x - h)^2} + {(y - k)^2} = {r^2}$
The above equation is known as the standard form for the equation of circle.
In this question ,we will first write the equation in standard form then, we will convert it in general form.

Complete step by step solution:
In this question , here h is 2 and k is 3
Let’s put the value of h and k in equation of circle for standard form of equation
${(x - 2)^2} + {(y - 3)^2} = {r^2}$
Now , it is given that the circle is tangent to x-axis.
Touching the x-axis means the distance from the centre is the radius.
$(2,3)$ represents $(x,y)$ coordinate that’s why radius is 3 units, therefore the equation will become,
${(x - 2)^2} + {(y - 3)^2} = {(3)^2}$
If you just solve the right hand side of this equation ${(x - 2)^2} + {(y - 3)^2} = {(3)^2}$ it will give you the standard form as ${(x - 2)^2} + {(y - 3)^2} = 9$
Now, solving the above equation to get the general form of equation of circle
Using ${(a - b)^2} = {a^2} + {b^2} - 2ab$ , simplify the above
$\Rightarrow {(x)^2} + {(2)^2} - 2(x)(2) + {(y)^2} + {(3)^2} - 2(y)(3)$
Simplifying the square
$\Rightarrow {x^2} + 4 - 4x + {y^2} + 9 - 6y = 9$
Combine like terms
$\Rightarrow {x^2} + {y^2} - 4x - 6y + 4 + 9 = 9$
Simplifying it further to get the general form
$\Rightarrow {x^2} + {y^2} - 4x - 6y + 4 = 0$

Thus, the general form of circle centred at $(2,3)$ and tangent to x-axis is ${x^2} + {y^2} - 4x - 6y + 4 = 0$

Note:

1. If you want to convert general form into standard form you can do it by completing the square method.
2. Radius is always positive.
3. Use brackets to avoid confusion in squaring and combining.