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 this circle, is :

Hyperbola PYQ — JEE Main 2019 · 11 Jan · Shift 2 · Q38 with Solution

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Correct answer: B

Interpret the given conditions

Let the centre of the circle be $(h,k)$ and radius be $r$ .

So the equation of the circle is

(x-h)^2+(y-k)^2=r^2.

The circle satisfies two conditions:

It cuts a chord of length $4a$ on the $x$ -axis.
It passes through a point on the $y$ -axis at distance $2b$ from the origin.

A point on the $y$ -axis at distance $2b$ from the origin is either $(0,2b)$ or $(0,-2b)$ . Since the locus will be the same form, we may take the point as $(0,2b)$ .

Use the chord-length condition on the x-axis

The distance of the centre $(h,k)$ from the $x$ -axis is $|k|$ .

If a circle of radius $r$ is cut by the $x$ -axis, then the length of the chord on the $x$ -axis is

2\sqrt{r^2-k^2}.

This is given as $4a$ , so

2\sqrt{r^2-k^2}=4a.

Thus,

\sqrt{r^2-k^2}=2a \quad\Rightarrow\quad r^2-k^2=4a^2.

Hence,

r^2=4a^2+k^2.

Use the condition that the circle passes through $(0,2b)$

Substitute $(0,2b)$ into the circle equation:

(0-h)^2+(2b-k)^2=r^2.

So,

h^2+(2b-k)^2=r^2.

Now replace $r^2$ by $4a^2+k^2$ :

h^2+(2b-k)^2=4a^2+k^2.

Expand:

h^2+4b^2-4bk+k^2=4a^2+k^2.

Cancelling $k^2$ from both sides,

h^2+4b^2-4bk=4a^2.

Rearrange:

h^2-4bk+4b^2-4a^2=0.

So,

h^2=4b(k+b)-4a^2.

Or,

4bk=h^2+4b^2-4a^2.

Hence,

k=\frac{h^2}{4b}+\frac{b^2-a^2}{b}.

Write the locus in standard coordinate form

If we denote the centre by $(x,y)$ instead of $(h,k)$ , then the locus is

y=\frac{x^2}{4b}+\frac{b^2-a^2}{b}.

This is of the form

y=Ax^2+C,

which is a parabola.

Check the options

A: ellipse — incorrect
B: parabola — correct
C: hyperbola — incorrect
D: straight line — incorrect

Therefore, the locus of the centre is a parabola.

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View written solutionFree

Correct answer: B

Interpret the given conditions

Let the centre of the circle be $(h,k)$ and radius be $r$ .

So the equation of the circle is

(x-h)^2+(y-k)^2=r^2.

The circle satisfies two conditions:

It cuts a chord of length $4a$ on the $x$ -axis.
It passes through a point on the $y$ -axis at distance $2b$ from the origin.

A point on the $y$ -axis at distance $2b$ from the origin is either $(0,2b)$ or $(0,-2b)$ . Since the locus will be the same form, we may take the point as $(0,2b)$ .

Use the chord-length condition on the x-axis

The distance of the centre $(h,k)$ from the $x$ -axis is $|k|$ .

If a circle of radius $r$ is cut by the $x$ -axis, then the length of the chord on the $x$ -axis is

2\sqrt{r^2-k^2}.

This is given as $4a$ , so

2\sqrt{r^2-k^2}=4a.

Thus,

\sqrt{r^2-k^2}=2a \quad\Rightarrow\quad r^2-k^2=4a^2.

Hence,

r^2=4a^2+k^2.

Use the condition that the circle passes through $(0,2b)$

Substitute $(0,2b)$ into the circle equation:

(0-h)^2+(2b-k)^2=r^2.

So,

h^2+(2b-k)^2=r^2.

Now replace $r^2$ by $4a^2+k^2$ :

h^2+(2b-k)^2=4a^2+k^2.

Expand:

h^2+4b^2-4bk+k^2=4a^2+k^2.

Cancelling $k^2$ from both sides,

h^2+4b^2-4bk=4a^2.

Rearrange:

h^2-4bk+4b^2-4a^2=0.

So,

h^2=4b(k+b)-4a^2.

Or,

4bk=h^2+4b^2-4a^2.

Hence,

k=\frac{h^2}{4b}+\frac{b^2-a^2}{b}.

Write the locus in standard coordinate form

If we denote the centre by $(x,y)$ instead of $(h,k)$ , then the locus is

y=\frac{x^2}{4b}+\frac{b^2-a^2}{b}.

This is of the form

y=Ax^2+C,

which is a parabola.

Check the options

A: ellipse — incorrect
B: parabola — correct
C: hyperbola — incorrect
D: straight line — incorrect

Therefore, the locus of the centre is a parabola.

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