Questions & Answers

Question

Answers

A. (3, 4)

B. (3, -4)

C. (8, -5)

D. (-8, 5)

Answer

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In the question, we are said that, square is drawn inside or inscribed in a circle with a given equation \[{{x}^{2}}+{{y}^{2}}-2x+4y-3=0\] with a given condition that is the sides of square parallel to the axis. For the given condition, we have to find one vertex of the square.

The given equation of circle is,

\[{{x}^{2}}+{{y}^{2}}-2x+4y-3=0\]

We will further write the equation as,

\[\begin{align}

& {{x}^{2}}-2x+1+{{y}^{2}}+4y+4-3-5=0 \\

& \Rightarrow {{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}-8=0 \\

\end{align}\]

So, the equation is formed as,

\[{{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=8={{\left( 2\sqrt{2} \right)}^{2}}\]

If the equation of circle is in form of,

\[{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}\]

Then, its center is \[\left( {{x}_{1}},{{y}_{1}} \right)\] and radius is r.

As the equation is \[{{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{\left( 2\sqrt{2} \right)}^{2}}\] so, its center is \[\left( 1,-2 \right)\] and radius is \[2\sqrt{2}\]

Let ABCD be the square inside the circle, whose center is O. So, it can be drawn as,

As we know that, the radius of the circle is \[2\sqrt{2}\] so, the diameter will be twice as radius \[4\sqrt{2}\].

The diameter of the circle is a diagonal of a square. Using the relation, diagonal of square \[\sqrt{2}\times \text{side of square}\] we can find the side of the square.

Here, side is AB, so, its length will be \[\dfrac{\text{diagonal}}{\sqrt{2}}\Rightarrow \dfrac{4\sqrt{2}}{\sqrt{2}}\Rightarrow 4\]

Now, let’s suppose, coordinates of B be (a, b).

Now, as we know that, each side length is 4 and sides are parallel to axes, we can write other coordinates in terms of a and b too.

So, the coordinates of A, C and D are \[\left( a-4,b \right);\left( a,b-4 \right);\left( a-4,b-4 \right)\].

We know that, midpoint of the diagonal of a square is the center of the circle. So, we can say that the midpoint of the coordinator of B and D is the center of the circle.

We will find midpoint using formula, \[x'=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}\,\,and\,\,y'=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\]

If $\left( x',y' \right)$ is the midpoint between \[\left( {{x}_{1}},{{y}_{1}} \right)\,\,and\,\,\left( {{x}_{2}},{{y}_{2}} \right)\] we can find midpoint using formula, \[x'=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}\,\,and\,\,y'=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\].

Here, points are B(a, b) and D(a-4, b-4), so, its midpoint will be,

\[\left( \dfrac{a+a-4}{2},\dfrac{b+b-4}{2} \right)\Rightarrow \left( \dfrac{2a-4}{2},\dfrac{2b-4}{2} \right)\]

Now, as we also know that its midpoint is the center of the circle which is (1, -2).

So we can say,

\[\left( \dfrac{\left( 2a-4 \right)}{2},\dfrac{\left( 2b-4 \right)}{2} \right)=\left( 1,-2 \right)\]

So, we can say that,

\[\begin{align}

& \dfrac{2a-4}{2}=1\,\,and\,\,\dfrac{2b-4}{2}=-2 \\

& \Rightarrow 2a-4=2\,\,and\,\,2b-4=-4 \\

\end{align}\]

Thus, on simplifying we can say that,

\[\begin{align}

& 2a=6\,\,\Rightarrow a=3 \\

& and\,\,2b=0\,\,\Rightarrow b=0 \\

\end{align}\]

So, the coordinates of B are (3, 0).

The coordinates of A is (-1, 0), C is (3, -4) and D is (-1, -4).

Among the above coordinates only (3, -4) matches the option.