# Conic Sections

## Class 11 NCERT

### NCERT

1   Find the equation of the circle with centre $(0, 2)$ and radius $2$

The equation of a circle with centre $(h, k)$ and radius $r$ is given as $(x - h) 2 + (y - k) 2 = r^ 2$$\\ It is given that centre (h, k) = (0, 2) and radius (r) = 2. Therefore, the equation of the circle is (x-0)^2+(y-2)^2 =2^2\\x^2+y^2-4-4y=4\\ x^2+y^2-4y=0 2 Find the equation of the circle with centre (-2,3) and radius 4 ##### Solution : The equation of a circle with centre (h, k) and radius r is given as (x - h) ^2 + (y - k) ^2 = r^ 2$$\\$ It is given that centre $(h, k) = (-2, 3)$ and radius $(r) = 4.$ Therefore, the equation of the circle is $(x + 2)^ 2 + (y - 3) ^2 = (4)^ 2\\ x^ 2 + 4x + 4 + y ^2 - 6y + 9 = 16\\ x^ 2 + y ^2 + 4x - 6y - 3 = 0$

3   Find the equation of the circle with centre $(\dfrac{1}{2},\dfrac{1}{4})$and radius $(\dfrac{1}{12})$

##### Solution :

The equation of a circle with centre $(h, k)$ and radius $r$ is given as$\\$ $(x-h)^2+(y-k)^2=r^2$$\\ It is given that centre (h, k) = (\dfrac{1}{2},\dfrac{1}{4}) and radius(r)=(\dfrac{1}{12})$$\\$ Therefore, the equation of the circle is $\\$ $(x-\dfrac{1}{2})^2+(y-\dfrac{1}{4})^2=(\dfrac{1}{12})^2\\ x^2-x+\dfrac{1}{4}+y^2-\dfrac{y}{2}+\dfrac{1}{16}=\dfrac{1}{144}\\ x^2-x+\dfrac{1}{4}+y^2-\dfrac{y}{2}+\dfrac{1}{16}-\dfrac{1}{144}=0\\ 144x^2-144x+36+144y^2-72y+9-1=0\\ 144x^2-144x+144y^2-72y+44=0\\ 36x^2-36x+36y^2-18y+11=0\\ 36x^2+36y^2-36x-18y+11=0$

4   Find the equation of the circle with centre $(1, 1)$ and radius $\sqrt{2}$

##### Solution :

The equation of a circle with centre $(h, k)$ and radius $r$ is given as $(x - h) ^2 + (y - k) ^2 = r2$$\\ It is given that centre (h, k) = (1, 1) and radius (r) = 2 .$$\\$ Therefore, the equation of the circle is$\\$ $(x-1)^2+(y-1)^2 =(\sqrt{2})^2\\ x^2-2x+1+y^2-2y+1=2\\ x^2+y^2-2x-2y=0$

5   Find the equation of the circle with centre $(-a, -b)$ and radius $\sqrt{a^2-b^2}$

##### Solution :

The equation of a circle with centre $(h, k)$ and radius $r$ is given as $(x - h)^ 2 + (y - k)^ 2 = r^ 2$$\\ It is given that centre (h, k) = (-a, -b) and radius (r) =\sqrt{a^2-b^2}.$$\\$ Therefore, the equation of the circle is$\\$ $(x+a)^2+(y+b)^2=(\sqrt{a^2-b^2})^2\\ x^2+2ax+a^2+y^2+2by+b^2\\ =a^2-b^2\\ x^2+y^2+2ax+2by+2b^2=0$