Show that the point $(x, y)$ given by $x = \dfrac{2at}{1+t^2}$ and $y = \dfrac{a(1-t^2)}{1+t^2}$ lies on a circle for all real values of $t$ such that $-1 \le t \le 1$ where $a$ is any given real numbers.

> [!TLDR] > Eliminate the parameter $t$ from the given equations to show that $x^2 + y^2 = a^2$, which is the equation of a circle centered at the origin with radius $|a|$. The parametric equations describe a point whose coordinates depend on a parameter $t$. To prove that all such points lie on a circle, we need to find a relationship between $x$ and $y$ that doesn't involve $t$ — and if that relationship turns out to be the equation of a circle, we're done. The standard form of a circle centered at the origin is $x^2 + y^2 = r^2$. Our strategy is to compute $x^2 + y^2$ using the given parametric forms and see what emerges. ## Finding $x^2 + y^2$ 1. **Square the expression for $x$:** $$x^2 = \left(\frac{2at}{1+t^2}\right)^2 = \frac{4a^2t^2}{(1+t^2)^2}$$ 2. **Square the expression for $y$:** $$y^2 = \left(\frac{a(1-t^2)}{1+t^2}\right)^2 = \frac{a^2(1-t^2)^2}{(1+t^2)^2}$$ 3. **Add $x^2$ and $y^2$:** Since both expressions have the same denominator $(1+t^2)^2$, we can combine them: $$x^2 + y^2 = \frac{4a^2t^2 + a^2(1-t^2)^2}{(1+t^2)^2}$$ 4. **Expand the numerator:** Focus on the numerator: $4a^2t^2 + a^2(1-t^2)^2$. First expand $(1-t^2)^2 = 1 - 2t^2 + t^4$. So the numerator becomes: $$4a^2t^2 + a^2(1 - 2t^2 + t^4) = 4a^2t^2 + a^2 - 2a^2t^2 + a^2t^4$$ $$= a^2 + 2a^2t^2 + a^2t^4$$ 5. **Factor the numerator:** Notice that $a^2 + 2a^2t^2 + a^2t^4 = a^2(1 + 2t^2 + t^4)$. The expression $1 + 2t^2 + t^4$ is a perfect square: $(1 + t^2)^2$. Therefore: $$x^2 + y^2 = \frac{a^2(1+t^2)^2}{(1+t^2)^2}$$ 6. **Simplify:** For all $t$ in the given range $-1 \le t \le 1$, we have $1 + t^2 > 0$, so $(1+t^2)^2 \ne 0$ and we can cancel: $$x^2 + y^2 = a^2$$ > [!FORMULA] > $$x^2 + y^2 = a^2$$ This is the equation of a circle centered at the origin with radius $|a|$. Since this relationship holds for all values of $t$ in the specified range (and in fact for all real $t$), every point $(x, y)$ generated by the parametric

Show that the point (x, y) given by x = (2at)/(1+t²) and y = (a(1-t²))/(1+t²) lies on a circle for all real va

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Concept understanding — Circle Equation Standard Form

Where the Circle Equation Comes From

Imagine you're standing at a point on a flat field. You tie a rope to a stake at that point, walk out the full length of the rope, and start walking in a circle, keeping the rope taut. Every point you step on is exactly the same distance from the stake.

That's the entire idea: a circle is the set of all points that are a fixed distance (the radius) from a fixed point (the centre).

If we put this on a coordinate plane, we can turn that geometric idea into an algebraic equation.

From Geometry to Algebra

Let the centre be at coordinates $(h, k)$ . Let the radius be $r$ . Take any point $(x, y)$ that lies on the circle. The distance from $(x, y)$ to $(h, k)$ must equal $r$ .

What's the distance between two points in the plane? The distance formula:

\sqrt{(x - h)^2 + (y - k)^2} = r

Now square both sides to remove the square root:

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

That's it. That's the standard form of the equation of a circle.

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

What Each Piece Tells You

$(x - h)$ and $(y - k)$ — these shift the circle away from the origin. If the centre is at $(0,0)$ , the equation simplifies to $x^2 + y^2 = r^2$ .
$r^2$ — notice it's the square of the radius, not the radius itself. If the equation says $x^2 + y^2 = 25$ , the radius is $\sqrt{25} = 5$ , not 25.
The equals sign — only the points $(x, y)$ that make this equation true lie on the circle. Any other point gives a larger or smaller left-hand side.

Watch out

A common mistake: for $(x - 3)^2 + (y + 2)^2 = 16$ , students often read the centre straight off the signs printed in the equation and say $(3, 2)$ . That's wrong. Each bracket must first be written in the exact form $x - h$ and $y - k$ : here $(y + 2) = (y - (-2))$ , so $k = -2$ , not $2$ . The centre is actually $(3, -2)$ . Always flip the sign inside every bracket before reading off $h$ and $k$ .

Quick Example

Write the equation of a circle with centre $(-1, 4)$ and radius $3$ .

Here $h = -1$ , $k = 4$ , $r = 3$ . Plug in:

(x - (-1))^2 + (y - 4)^2 = 3^2

Simplify:

(x + 1)^2 + (y - 4)^2 = 9

That's the standard form. From this, you can immediately read off the centre $(-1, 4)$ and radius $3$ .

Why This Form Matters

The standard form is the most useful because it gives you the centre and radius at a glance. In exams, you'll often be given an expanded form like $x^2 + y^2 - 6x + 4y - 12 = 0$ and asked to rewrite it in standard form by completing the square — that's the next step in your learning, but the standard form itself is the destination.

For now: centre tells you where, radius tells you how big, and the equation tells you which points belong.

The Standard Form of a Circle's Equation is one of the first results in the NCERT Class 11 Mathematics chapter on Conic Sections, matching searches like "equation of a circle: definition, formula and examples" or "conic sections important questions class 11 maths". Recognising centre and radius directly from this form is also a routine, quick-scoring question type in CBSE boards, JEE Main, and state CET coordinate geometry sections.

Eliminate the parameter $t$ from the given equations to show that $x^2 + y^2 = a^2$ , which is the equation of a circle centered at the origin with radius $|a|$ .

The parametric equations describe a point whose coordinates depend on a parameter $t$ . To prove that all such points lie on a circle, we need to find a relationship between $x$ and $y$ that doesn't involve $t$ — and if that relationship turns out to be the equation of a circle, we're done.

The standard form of a circle centered at the origin is $x^2 + y^2 = r^2$ . Our strategy is to compute $x^2 + y^2$ using the given parametric forms and see what emerges.

Finding $x^2 + y^2$

Square the expression for $x$ :

x^2 = \left(\frac{2at}{1+t^2}\right)^2 = \frac{4a^2t^2}{(1+t^2)^2}

Square the expression for $y$ :

y^2 = \left(\frac{a(1-t^2)}{1+t^2}\right)^2 = \frac{a^2(1-t^2)^2}{(1+t^2)^2}

Add $x^2$ and $y^2$ : Since both expressions have the same denominator $(1+t^2)^2$ , we can combine them:

x^2 + y^2 = \frac{4a^2t^2 + a^2(1-t^2)^2}{(1+t^2)^2}

Expand the numerator:

Focus on the numerator: $4a^2t^2 + a^2(1-t^2)^2$ .

First expand $(1-t^2)^2 = 1 - 2t^2 + t^4$ .

So the numerator becomes:

4a^2t^2 + a^2(1 - 2t^2 + t^4) = 4a^2t^2 + a^2 - 2a^2t^2 + a^2t^4

= a^2 + 2a^2t^2 + a^2t^4

Factor the numerator:

Notice that $a^2 + 2a^2t^2 + a^2t^4 = a^2(1 + 2t^2 + t^4)$ .

The expression $1 + 2t^2 + t^4$ is a perfect square: $(1 + t^2)^2$ .

Therefore:

x^2 + y^2 = \frac{a^2(1+t^2)^2}{(1+t^2)^2}

Simplify: For all $t$ in the given range $-1 \le t \le 1$ , we have $1 + t^2 > 0$ , so $(1+t^2)^2 \ne 0$ and we can cancel:

x^2 + y^2 = a^2

x^2 + y^2 = a^2

This is the equation of a circle centered at the origin with radius $|a|$ . Since this relationship holds for all values of $t$ in the specified range (and in fact for all real $t$ ), every point $(x, y)$ generated by the parametric equations lies on this circle.

Tip

The parametric form given here is actually a rational parametrization of the circle, closely related to the Weierstrass substitution used in trigonometric integration. If you set $t = \tan(\theta/2)$ , these formulas become $x = a\sin\theta$ and $y = a\cos\theta$ .

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Q.Show that the point $(x, y)$ given by $x = \dfrac{2at}{1+t^2}$ and $y = \dfrac{a(1-t^2)}{1+t^2}$ lies on a circle for all real values of $t$ such that $-1 \le t \le 1$ where $a$ is any given real numbers.

Finding $x^2 + y^2$

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