CBSE Class 11 Mathematics Chapter 11: Conic Sections NCERT Solutions

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

In this problem, we are given the centre of the circle as $(h,\; k)\; =\; (0,\; 2)$ and the radius as $r\; =\; 2$.

Substituting these values into the standard equation, we get:

$(x-0)^2\; +\; (y-2)^2\; =\; 2^2$

Expanding the terms:

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

Simplifying the equation by subtracting 4 from both sides:

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

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

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

We are given that the centre of the circle is $(h,\; k)\; =\; (-2,\; 3)$ and the radius is $r\; =\; 4$.

Substitute these values into the standard equation:

$(x\; -\; (-2))^2\; +\; (y\; -\; 3)^2\; =\; 4^2$

$(x\; +\; 2)^2\; +\; (y\; -\; 3)^2\; =\; 16$

Now, expand the squared terms:

$(x^2\; +\; 4x\; +\; 4)\; +\; (y^2\; -\; 6y\; +\; 9)\; =\; 16$

Combine the constant terms and rearrange the equation:

$x^2\; +\; y^2\; +\; 4x\; -\; 6y\; +\; 13\; =\; 16$

Subtract 16 from both sides to set the equation to zero:

$$x^2\; +\; y^2\; +\; 4x\; -\; 6y\; -\; 3\; =\; 0$$

The equation of the circle is $x^2\; +\; y^2\; +\; 4x\; -\; 6y\; -\; 3\; =\; 0$.

The standard form of a circle's equation is $(x-h)^2\; +\; (y-k)^2\; =\; r^2$, where $(h,\; k)$ is the centre and $r$ is the radius.

Given centre $(h,\; k)\; =\; \backslash left(\backslash frac\{1\}\{2\},\; \backslash frac\{1\}\{4\}\backslash right)$ and radius $r\; =\; \backslash frac\{1\}\{12\}$.

Substitute these values into the standard equation:

$\backslash left(x\; -\; \backslash frac\{1\}\{2\}\backslash right)^2\; +\; \backslash left(y\; -\; \backslash frac\{1\}\{4\}\backslash right)^2\; =\; \backslash left(\backslash frac\{1\}\{12\}\backslash right)^2$

Expand the terms:

$x^2\; -\; 2\; \backslash cdot\; x\; \backslash cdot\; \backslash frac\{1\}\{2\}\; +\; \backslash left(\backslash frac\{1\}\{2\}\backslash right)^2\; +\; y^2\; -\; 2\; \backslash cdot\; y\; \backslash cdot\; \backslash frac\{1\}\{4\}\; +\; \backslash left(\backslash frac\{1\}\{4\}\backslash right)^2\; =\; \backslash frac\{1\}\{144\}$

$x^2\; -\; x\; +\; \backslash frac\{1\}\{4\}\; +\; y^2\; -\; \backslash frac\{y\}\{2\}\; +\; \backslash frac\{1\}\{16\}\; =\; \backslash frac\{1\}\{144\}$

To eliminate fractions, find a common denominator, which is 144. Multiply every term by 144:

$144(x^2)\; -\; 144(x)\; +\; 144\backslash left(\backslash frac\{1\}\{4\}\backslash right)\; +\; 144(y^2)\; -\; 144\backslash left(\backslash frac\{y\}\{2\}\backslash right)\; +\; 144\backslash left(\backslash frac\{1\}\{16\}\backslash right)\; =\; 144\backslash left(\backslash frac\{1\}\{144\}\backslash right)$

$144x^2\; -\; 144x\; +\; 36\; +\; 144y^2\; -\; 72y\; +\; 9\; =\; 1$

Rearrange the terms and set the equation to zero:

$144x^2\; +\; 144y^2\; -\; 144x\; -\; 72y\; +\; 36\; +\; 9\; -\; 1\; =\; 0$

$144x^2\; +\; 144y^2\; -\; 144x\; -\; 72y\; +\; 44\; =\; 0$

Divide the entire equation by the greatest common divisor, which is 4, to simplify:

$36x^2\; +\; 36y^2\; -\; 36x\; -\; 18y\; +\; 11\; =\; 0$

The equation of the circle is $36x^2\; +\; 36y^2\; -\; 36x\; -\; 18y\; +\; 11\; =\; 0$.

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

We are given the centre $(h,\; k)\; =\; (1,\; 1)$ and the radius $r\; =\; \backslash sqrt\{2\}$.

Substitute these values into the standard equation:

$(x-1)^2\; +\; (y-1)^2\; =\; (\backslash sqrt\{2\})^2$

Expand the squared terms:

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

Combine the terms:

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

Subtract 2 from both sides to simplify the equation:

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

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