Answer

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**Hint:**Here, we will first plot a graph using the details given in the question. We will use the fact that a line drawn from the centre of a circle to a chord bisects the chord. Then, we will apply the Pythagoras theorem and with the help of the equation of a circle we would substitute the values and while eliminating, reach a step where we would get the required locus of the centre of the circle.

**Formulas Used:**

We will use the following formulas:

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

2. \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]

**Complete step-by-step answer:**

For solving this question, firstly, we will draw a circle on a graph such that the two points where the circle intersects the X-axis makes a length of \[4a\].

Distance PQ \[ = 4a\] (from the graph shown below)

Any of the two points (whether above or below the origin) where the circle intersects the Y-axis could be taken, where its length is \[2b\] from the origin.

i.e. Coordinates of point R \[ = \left( {0,2b} \right)\] (from the graph shown below)

Now, let the coordinates of the centre of the circle O \[ = \left( {h,k} \right)\].

Also, PQ is the required chord on the X-axis on which the circle makes a length of \[4a\] .

When we draw a perpendicular from the centre of the circle to the chord PQ,

Then, we know that, a perpendicular drawn from the centre of a circle to a chord, bisects the chord

i.e. it divides the chord in 2 equal lengths

\[ \Rightarrow \] PS \[=\] SQ \[= \dfrac{{4a}}{2} = 2a\]

Also, OS \[= k\] (because the \[\left( {x,y} \right)\] coordinates of centre O is \[\left( {h,k} \right)\] respectively.

And, OQ \[=\] radius of the circle \[ = r\]

Now, applying Pythagoras theorem in the triangle OSQ where, OQ is the hypotenuse, we get

\[{\left( {{\rm{OQ}}} \right)^{2}} = {\left( {{\rm{OS}}} \right)^2} + {\left( {{\rm{SQ}}} \right)^2}\]

\[ \Rightarrow {r^2} = {k^2} + {\left( {2a} \right)^2}\]

Solving further, we get,

\[{r^2} = {k^2} + 4{a^2}\]………………………….\[\left( 1 \right)\]

_{}

Now, equation of a circle is \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]

Where, \[\left( {h,k} \right)\] are the coordinates of the centre of the circle and \[r\] is the radius.

Also, \[x\] and \[y\] are the coordinates of any generic point lying on the circle.

Now, knowing the coordinates of point R and substituting \[x\]and \[y\] by \[0\] and \[2b\] respectively, we get equation of given circle as

\[{\left( {0 - h} \right)^2} + {\left( {2b - k} \right)^2} = {r^2}\]

\[ \Rightarrow {h^2} + {\left( {2b - k} \right)^2} = {r^2}\]

Substituting the value of \[{r^2}\] from \[\left( 1 \right)\] we get,

\[{h^2} + {\left( {2b - k} \right)^2} = {k^2} + 4{a^2}\]

Simplifying the equation using the formula

\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]

\[ \Rightarrow {h^2} + 4{b^2} + {k^2} - 4bk = {k^2} + 4{a^2}\]

Eliminating \[{k^2}\] on both the sides and shifting some variables to R.H.S. we get,

\[ \Rightarrow {h^2} = 4{a^2} + 4bk - 4{b^2}\]

Here, \[h\] and \[k\] represents \[x\] and \[y\] coordinates respectively,

Hence, by replacing them, we get,

\[{x^2} = 4{a^2} + 4by - 4{b^2}\]

Now, if we observe the LHS and RHS then,

It is likely in the form:

\[{x^2} = 4ay\]

Clearly, this equation is nearest to the equation of Parabola.

Hence, the locus of the centre of the circle is a parabola.

**Therefore, option B is the correct option.**

**Note:**

There is another way to solve this question.

As we know, the General Form of the equation of a circle is \[{x^2} + {y^2} + 2gx + 2fy + c = 0\]

Where,

The centre of the circle is \[\left( { - g, - f} \right)\].

The radius of the circle is \[\sqrt {{g^2} + {f^2} - c} \].

If we are given a circle in the general form then we can change it into the standard form by completing the square.

Now, equation of circle will be

\[{x^2} + {y^2} + 2gx + 2fy + c = 0\]

Also, knowing the coordinates of point R and substituting \[x\] and \[y\] by \[0\] and \[2b\] respectively, we get ,

Equation of circle =

\[{\left( 0 \right)^2} + {\left( {2b} \right)^2} + 2g\left( 0 \right) + 2f\left( 2 \right) + c = 0\]

Solving further,

\[ \Rightarrow 4{b^2} + 4f + c = 0\]……………………………\[\left( 1 \right)\]

Now, we know that

\[2\sqrt {{g^2} - c} = 4a\]

Squaring both sides,

\[ \Rightarrow {g^2} - c = \dfrac{{16{a^2}}}{4} = 4{a^2}\]

\[ \Rightarrow c = {g^2} - 4{a^2}\]

Putting the value of \[c\] in equation \[\left( 1 \right)\] , we get,

\[4{b^2} + 4f + {g^2} - 4{a^2} = 0\]

Solving further,

\[ \Rightarrow 4\left( {{b^2} - {a^2}} \right) + 4f + {g^2} = 0\]

Finally, substituting \[g\] and \[f\] by \[x\] and \[y\] respectively, we get,

\[ \Rightarrow {x^2} + 4y + 4\left( {{b^2} - {a^2}} \right) = 0\]

This represents a Parabola.

Hence, option B is the correct option.

This method could be applied if we know the general equation of a circle as this method would take less time than the previous one.

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