Question

How do you find the equation of the circle given centre at the $(10,5)$ and a radius of $11$ ?

Answer 1

Hint: In the question above, we have the radius of the circle and also, the centre points. In order to solve this question, we need to know the formula of finding the equation of a circle with the given values. Also, we have the standard form of the equation of circle which is:
${(x - a)^2} + {(y - b)^2} = {r^2}$ where $(a,b)$ are the coordinates of the centre and $r$ is the radius of the circle.

Complete step-by-step solution:
Here, we have a circle with a centre point $(10,5)$ and a radius of $11$ .
Now, when finding the equation of a circle, we must first know the required formulae and the needed values in order to substitute them in the right positions.
We all are aware of the standard form of the equation of a circle:
${(x - a)^2} + {(y - b)^2} = {r^2}$
Where $(a,b)$ are the coordinates of the centre.
Here $(a,b)$ = $(10,5)$ and $r = 11$
Putting the values in the formula,
$ \Rightarrow {(x - 10)^2} + {(y - 5)^2} = 121$ , since we know the radius.
Solving the brackets with the ${(a - b)^2} = {a^2} - 2ab + {b^2}$ formula,
$ \Rightarrow ({x^2} - 20x + 100) + ({y^2} - 10y + 25) = 121$
Taking out the brackets as per the requirement of changing signs,
$ \Rightarrow {x^2} - 20x + 100 + {y^2} - 10y + 25 = 121$
Solving the given numerical equation further,
$ \Rightarrow {x^2} - 20x + {y^2} - 10y + 125 = 121$
Shifting the constants to subtract it,
$ \Rightarrow {x^2} - 20x + {y^2} - 10y + 125 - 121 = 0$
Simplifying further,
$ \Rightarrow {x^2} + {y^2} - 20x - 10y + 4 = 0$

Therefore, for a circle with centre as $(10,5)$ and radius as $11$ , the equation will be written as ${x^2} + {y^2} - 20x - 10y + 4 = 0$

Note: If the equation of a circle is in the standard form, it helps us to find the centre of the circle, and also the radius of the same, very easily. The radius of a circle is always positive. The equation of a circle comes in two forms:
The standard form: ${(x - a)^2} + {(y - b)^2} = {r^2}$
The general form: ${x^2} + {y^2} + Dx + Ey + F = 0$ where $D,E,F$ are constants.