Question

Find the equation of the circle in the first quadrant which touches each axis at a distance $5$ from the origin.
A. ${x^2} + {y^2} + 5x + 5y + 25 = 0$
B. ${x^2} + {y^2} - 10x - 10y + 25 = 0$
C. ${x^2} + {y^2} - 5x - 5y + 25 = 0$
D. ${x^2} + {y^2} + 10x + 10y + 25 = 0$

Answer 1

Hint: From the given data we will be drawing a figure which will help us to know the radius and the coordinates of the center from which we can substitute the values in the standard form of a circle.

Formula Used:
We will be using standard form of circle which is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$and we will be also using the formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ to expand the bracket.

Complete step by step solution:
We will be drawing the figure to understand the terms written in the question,

Image: Circle at a distance of 5 units.
According to the figure we can say that the centre of the circle is $\left( {h,k} \right)$ which is equal to,
$\left( {5,5} \right)$ and the radius is equal to $r = 5$
By using the standard form of circle we can substitute the values in it,
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$
$ \Rightarrow {\left( {x - 5} \right)^2} + {\left( {y - 5} \right)^2} = {5^2}$
Using the formula of ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
We will open the bracket in the equation.
$ \Rightarrow {\left( {x - 5} \right)^2} + {\left( {y - 5} \right)^2} = {5^2}$
$ \Rightarrow {x^2} - 10x + 25 + {y^2} - 10y + 25 = 25$
$ \Rightarrow {x^2} + {y^2} - 10x - 10y + 25 = 0$

Option ‘B’ is correct

Note: While drawing the figure student should be very much aware about the distance and the coordinates which will be taken in respect to expand the general form of the circle.