Question

ABCD is a square with side 'a'. If AB and AD are taken as positive coordinate axis then equation of circle circumscribing the square is

& A.{{x}^{2}}+{{y}^{2}}-ax-ay=0 \\\ & B.{{x}^{2}}+{{y}^{2}}+ax+ay=0 \\\ & C.{{x}^{2}}+{{y}^{2}}-ax+ay=0 \\\ & D.{{x}^{2}}+{{y}^{2}}+ax-ay=0 \\\ \end{aligned}$$

Answer 1

Hint: We will first draw a diagram to understand the question better. Then, we will find the center and radius of the circle and put them in the general equation of the circle to get the required equation of the circle. We will use Pythagora's theorem which states that, in a right angled triangle, the squares of the sum of base and perpendicular is equal to the square of hypotenuse. We will also use general equation of circle which is given by ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ where (h, k) represent center of circles and r represent radius of circle.

Complete step-by-step answer:

Let us first draw diagram for a better understanding of the question:

From the properties of the circle, we know that the perpendicular from the center of the circle to a chord bisect the chord. Here, in the diagram, side AD represents the chord of the circle. Therefore, a perpendicular line drawn from the center of circle to the square of side 'a' will bisect the side AD.

Similarly, AB represents chord and perpendicular drawn from center of circle to the side AB will bisect the side AB.

Now, if AD and AB are coordinate axes, half of 'a' will represent the coordinate points of the center. Hence, the center will be $\left( \dfrac{a}{2},\dfrac{a}{2} \right)$.

Let us join OA, we get a right angled triangle OAP, where P is a bisecting point on side AD. Using Pythagoras theorem,

${{\left( OA \right)}^{2}}={{\left( AP \right)}^{2}}+{{\left( OP \right)}^{2}}$

Since, AP is half of AD, therefore $AP=\dfrac{a}{2}$ and also OP is half of AB, therefore, $OP=\dfrac{a}{2}$ we get:

${{\left( OA \right)}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}}+{{\left( \dfrac{a}{2} \right)}^{2}}$

$\Rightarrow OA=\sqrt{\dfrac{2{{a}^{2}}}{4}}$

$\Rightarrow OA=\dfrac{a}{\sqrt{2}}$

As we can see, OA represents the radius of a circle. Hence, radius of circle is $\dfrac{a}{\sqrt{2}}$

General equation of circle is given by ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ where (h, k) represent center of circle and r represents radius of circle. Putting values we get:

${{\left( x-\left( \dfrac{a}{2} \right) \right)}^{2}}+{{\left( y-\left( \dfrac{a}{2} \right) \right)}^{2}}={{\left( \dfrac{a}{\sqrt{2}} \right)}^{2}}$

$\Rightarrow {{x}^{2}}-ax+\dfrac{{{a}^{2}}}{4}+{{y}^{2}}-ay+\dfrac{{{a}^{2}}}{4}=\dfrac{{{a}^{2}}}{2}$

$\Rightarrow {{x}^{2}}+{{y}^{2}}-ax-ay+\dfrac{{{a}^{2}}}{2}=\dfrac{{{a}^{2}}}{2}$

$\Rightarrow {{x}^{2}}+{{y}^{2}}-ax-ay=0$

${{x}^{2}}+{{y}^{2}}-ax-ay=0$ which is the required equation of the circle.

So, the correct answer is “Option A”.

Note: Students should carefully calculate center and radius of the circle. Care must be taken while using Pythagoras theorem. At least, students should take care of plus minus signs which is a very common mistake in these types of questions.