# 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$\\$ $(x+a)^2+(y+b)^2=(\sqrt{a^2-b^2})^2\\ x^2+2ax+a^2+y^2+2by+b^2\\ =a^2-b^2\\ x^2+y^2+2ax+2by+2b^2=0$

6   Find the centre and radius of the circle $(x + 5) ^2 + (y - 3)^ 2 = 36$

##### Solution :

The equation of the given circle is $(x + 5)^ 2 + (y - 3)^ 2 = 36.$$\\ (x + 5) 2 + (y - 3) 2 = 36$$\\$ $\Rightarrow \{x - (-5)\} ^2 + (y - 3) ^2 = 6 ^2 ,$ which is of the form $(x - h) ^2 + (y - k)^ 2 = r _2$ , where $h = -5, k = 3,$ and $r = 6$.$\\$ Thus, the centre of the given circle is $(-5, 3)$, while its radius is $6$.

7   Find the centre and radius of the circle $x ^2 + y^ 2 - 4x - 8y - 45 = 0$

##### Solution :

The equation of the given circle is $x^ 2 + y^ 2 - 4x - 8y - 45 = 0.$$\\ x^ 2 + y ^2 - 4x - 8y - 45 = 0$$\\$ $\Rightarrow (x^ 2 - 4x) + (y^ 2 - 8y) = 45$$\\ \Rightarrow {x ^2 - 2(x)(2) + 2 ^2 } + {y^ 2 - 2(y)(4)+ 4 ^2 } - 4 -16 = 45$$\\$ $\Rightarrow (x - 2) ^2 + (y -4) ^2 = 65$$\\ \Rightarrow (x - 2) ^2 + (y -4) ^2 =(\sqrt{65})^2$$\\$ , which is of the form $(x - h) ^2 + (y - k)^ 2 = r^ 2$ , where $h = 2, k = 4,$ and $r = \sqrt{65 }$.$\\$ Thus, the centre of the given circle is $(2, 4)$, while its radius is $\sqrt{65}$

8   Find the centre and radius of the circle $x^ 2 + y ^2 - 8x + 10y - 12 = 0$

The equation of the given circle is $x^ 2 + y ^2 - 8x + 10y - 12 = 0.$$\\ x^ 2 + y^ 2 - 8x + 10y - 12 = 0\\ \Rightarrow (x ^2 - 8x) + (y ^2 + 10y) = 12\\ \Rightarrow \{x ^2 - 2(x)(4) + 4 ^2 \} + \{y ^2 + 2(y)(5) + 5 ^2 \}- 16 - 25 = 12\\ \Rightarrow (x - 4) ^2 + (y + 5)^ 2 = 53\\ \Rightarrow (x-4)^2+\{y-(-5)\}^2=(\sqrt{53})^2 , which is of the form (x - h)^2 + (y - k) ^2 = r^ 2 , where h = 4,$$\\$ $k = -5$, and $r =\sqrt{53} .$$\\ Thus, the centre of the given circle is (4, -5), while its radius is \sqrt{53} 9 Find the centre and radius of the circle 2x^ 2 + 2y ^2 - x = 0 ##### Solution : The equation of the given circle is 2x ^2 + 2y ^2 - x = 0. 2x ^2 + 2y ^2 - x = 0.\\ \Rightarrow (2x^2-x)+2y^2=0\\ \Rightarrow 2[(x^2-\dfrac{x}{2})+y^2]=0\\ \Rightarrow \left\{x^2 -2. x\left(\dfrac{1}{4}\right)+\left(\dfrac{1}{4}\right)^2 \right\} +y^2-(\dfrac{1}{4})^2=0\\ \Rightarrow (x-\dfrac{1}{4})^2+(y-0)^2=(\dfrac{1}{4})^2, which is of the form (x - h)^ 2 + (y - k) ^2 = r^ 2 , where h =\dfrac{1}{4},k=0, and r=\dfrac{1}{4}$$\\$ Thus, the centre of the given circle is $(\dfrac{1}{4},0)$,while its radius is $\dfrac{1}{4}$

10   Find the equation of the circle passing through the points $(4, 1)$ and $(6, 5)$ and whose centre is on the line $4x + y = 16.$

##### Solution :

Let the equation of the required circle be $(x - h)^ 2 + (y - k)^ 2 = r^ 2 .$$\\ Since the circle passes through points (4, 1) and (6, 5),$$\\$ $(4 - h)^ 2 + (1 - k) ^2 = r^ 2 ... (1)$$\\ (6 - h)^2 + (5 - k)^2 = r^2 ... (2)$$\\$ Since the centre $(h, k)$ of the circle lies on line $4x + y = 16, 4h + k = 16 ... (3)$$\\ From equations (1) and (2), we obtain\\ (4 - h) ^2 + (1 - k)^ 2 = (6 - h) ^2 + (5 - k) ^2\\ \Rightarrow 16 -8h + h ^2 + 1 - 2k + k^ 2 = 36 - 12h + h^2 + 25 - 10k + k^ 2\\ \Rightarrow 16 - 8h + 1 - 2k = 36 - 12h + 25 - 10k\\ \Rightarrow 4h + 8k = 44\\ \Rightarrow h + 2k = 11 ... (4)$$\\$ On solving equations (3) and (4), we obtain $h = 3$ and $k = 4.$$\\ On substituting the values of h and k in equation (1), we obtain\\ (4 - 3) ^2 + (1 - 4) ^2 = r^ 2\\ \Rightarrow (1) ^2 + (- 3) ^2 = r^2\\ \Rightarrow 1 + 9 = r^2\\ \Rightarrow r^ 2 = 10\\ \Rightarrow r = 10$$\\$ Thus, the equation of the required circle is $(x - 3)^ 2 + (y - 4)^ 2 = (\sqrt{10 })^ 2\\ X^ 2 - 6x + 9 + y^ 2 -8y + 16 = 10\\ X^ 2 + y ^2 - 6x - 8y + 15 = 0$$\\$