CBSE Class 11 Maths Exemplar Chapter 11: Conic Sections NCERT Solutions

Given that the radius of the circle is 'a'. Since the circle touches both axes in the first quadrant, its center must be equidistant from both the x-axis and the y-axis. Therefore, the coordinates of the center (h, k) are (a, a).

The standard equation of a circle with center (h, k) and radius r is given by $$(x-h)^2\; +\; (y-k)^2\; =\; r^2$$.

Substituting the center (a, a) and radius a, we get:

$$(x-a)^2\; +\; (y-a)^2\; =\; a^2$$

Expanding this equation:

$$x^2\; -\; 2ax\; +\; a^2\; +\; y^2\; -\; 2ay\; +\; a^2\; =\; a^2$$

Simplifying the equation by cancelling one $a^2$ from both sides and rearranging terms:

$$x^2\; +\; y^2\; -\; 2ax\; -\; 2ay\; +\; a^2\; =\; 0$$

Thus, the equation of the required circle is $x^2\; +\; y^2\; -\; 2ax\; -\; 2ay\; +\; a^2\; =\; 0$.

We are given the parametric equations for a point (x, y) as:

$x\; =\; \backslash frac\{2at\}\{1\; +\; t^2\}$

$y\; =\; \backslash frac\{a(1\; -\; t^2)\}\{1\; +\; t^2\}$

To show that this point lies on a circle, we need to find a relationship between x and y that represents the equation of a circle. Let's try to eliminate the parameter 't'.

First, let's calculate $x^2$ and $y^2$:

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

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

Now, let's find the sum $x^2\; +\; y^2$:

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

Factor out $\backslash frac\{a^2\}\{(1\; +\; t^2)^2\}$:

$$x^2\; +\; y^2\; =\; \backslash frac\{a^2\}\{(1\; +\; t^2)^2\}\; \backslash left[\; 4t^2\; +\; (1\; -\; t^2)^2\; \backslash right]$$

Expand $(1\; -\; t^2)^2$:

$(1\; -\; t^2)^2\; =\; 1\; -\; 2t^2\; +\; t^4$

Substitute this back into the expression:

$$x^2\; +\; y^2\; =\; \backslash frac\{a^2\}\{(1\; +\; t^2)^2\}\; \backslash left[\; 4t^2\; +\; 1\; -\; 2t^2\; +\; t^4\; \backslash right]$$

Combine like terms inside the bracket:

$$x^2\; +\; y^2\; =\; \backslash frac\{a^2\}\{(1\; +\; t^2)^2\}\; \backslash left[\; t^4\; +\; 2t^2\; +\; 1\; \backslash right]$$

Recognize that $t^4\; +\; 2t^2\; +\; 1$ is a perfect square: $(t^2\; +\; 1)^2$.

$$x^2\; +\; y^2\; =\; \backslash frac\{a^2\}\{(1\; +\; t^2)^2\}\; \backslash left[\; (1\; +\; t^2)^2\; \backslash right]$$

Cancel out the $(1\; +\; t^2)^2$ terms:

$$x^2\; +\; y^2\; =\; a^2$$

This equation, $x^2\; +\; y^2\; =\; a^2$, is the standard equation of a circle centered at the origin (0, 0) with radius 'a'. Therefore, the given point (x, y) lies on this circle.

Let the general equation of a circle be $x^2\; +\; y^2\; +\; 2gx\; +\; 2fy\; +\; c\; =\; 0$. The center of this circle is at $(-g,\; -f)$.

Since the circle passes through the origin (0, 0), substituting these coordinates into the general equation gives:

$0^2\; +\; 0^2\; +\; 2g(0)\; +\; 2f(0)\; +\; c\; =\; 0\; \backslash Rightarrow\; c\; =\; 0$.

So, the equation of the circle simplifies to $x^2\; +\; y^2\; +\; 2gx\; +\; 2fy\; =\; 0$.

Now, the circle passes through the point (a, 0). Substituting these coordinates:

$a^2\; +\; 0^2\; +\; 2g(a)\; +\; 2f(0)\; =\; 0$

$a^2\; +\; 2ag\; =\; 0$

Since 'a' is a coordinate and likely non-zero for a distinct point, we can divide by 'a':

$a\; +\; 2g\; =\; 0\; \backslash Rightarrow\; g\; =\; -\backslash frac\{a\}\{2\}$

Next, the circle passes through the point (0, b). Substituting these coordinates:

$0^2\; +\; b^2\; +\; 2g(0)\; +\; 2f(b)\; =\; 0$

$b^2\; +\; 2bf\; =\; 0$

Assuming 'b' is non-zero:

$b\; +\; 2f\; =\; 0\; \backslash Rightarrow\; f\; =\; -\backslash frac\{b\}\{2\}$

The coordinates of the center of the circle are $(-g,\; -f)$.

Substituting the values of g and f we found:

Center = $\backslash left(-\backslash left(-\backslash frac\{a\}\{2\}\backslash right),\; -\backslash left(-\backslash frac\{b\}\{2\}\backslash right)\backslash right)\; =\; \backslash left(\backslash frac\{a\}\{2\},\; \backslash frac\{b\}\{2\}\backslash right)$

Therefore, the coordinates of the center of the circle are $\backslash left(\backslash frac\{a\}\{2\},\; \backslash frac\{b\}\{2\}\backslash right)$.

We are given that the center of the circle is C(1, 2).

The condition that the circle touches the X-axis means that the distance from the center of the circle to the X-axis is equal to the radius of the circle.

The distance from a point (h, k) to the X-axis is the absolute value of the y-coordinate, i.e., |k|.

In this case, the center is (1, 2), so h = 1 and k = 2.

The radius (r) of the circle is therefore $|2|\; =\; 2$.

The standard equation of a circle with center (h, k) and radius r is $(x-h)^2\; +\; (y-k)^2\; =\; r^2$.

Substituting the center (1, 2) and radius r = 2:

$$(x-1)^2\; +\; (y-2)^2\; =\; 2^2$$

Expanding the equation:

$$x^2\; -\; 2(1)x\; +\; 1^2\; +\; y^2\; -\; 2(2)y\; +\; 2^2\; =\; 4$$

$$x^2\; -\; 2x\; +\; 1\; +\; y^2\; -\; 4y\; +\; 4\; =\; 4$$

Simplify the equation by cancelling the '4' on both sides and rearranging terms:

$$x^2\; -\; 2x\; +\; 1\; +\; y^2\; -\; 4y\; =\; 0$$

$$x^2\; +\; y^2\; -\; 2x\; -\; 4y\; +\; 1\; =\; 0$$

Thus, the equation of the circle is $x^2\; +\; y^2\; -\; 2x\; -\; 4y\; +\; 1\; =\; 0$.