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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 is
A. A hyperbola
B. A parabola
C. A straight line
D. An ellipse

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Last updated date: 26th Feb 2024
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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,
\[{r^2} = {k^2} + 4{a^2}\]………………………….\[\left( 1 \right)\]
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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.