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# A circle cuts a chord of length $4a$ on the x-axis and passes through a point on the y-axis, distant $2b$ from the origin. Then the locus of the centre of the circle isA. A hyperbolaB. A parabolaC. A straight lineD. An ellipse

Last updated date: 24th Jul 2024
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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. ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$

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.