NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

Ans : 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
\(\begin{array}{l}{(x-0)^{2}+(y-2)^{2}=2^{2}} \\ {x^{2}+y^{2}+4-4 y=4} \\ {x^{2}+y^{2}-4 y=0}\end{array}\) 

Ans : 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 
\(\begin{array}{l}{(x+2)^{2}+(y-3)^{2}=(4)^{2}} \\ {x^{2}+4 x+4+y^{2}-6 y+9=16} \\ {x^{2}+y^{2}+4 x-6 y-3=0}\end{array}\) 

Ans : 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 \((\mathrm{h}, \mathrm{k})=\left(\frac{1}{2}, \frac{1}{4}\right){\text { and radius }(\mathrm{r})}=\frac{1}{12}\)
Therefore, the equation of the circle is
\(\begin{array}{l}{\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{4}\right)^{2}=\left(\frac{1}{12}\right)^{2}} \\ {x^{2}-x+\frac{1}{4}+y^{2}-\frac{y}{2}+\frac{1}{16}=\frac{1}{144}} \\ {x^{2}-x+\frac{1}{4}+y^{2}-\frac{y}{2}+\frac{1}{16}-\frac{1}{144}=0} \\ {144 x^{2}-144 x+36+144 y^{2}-72 y+9-1=0} \\ {144 x^{2}-36 x+144 y^{2}-18 y+41=0} \\ {36 x^{2}-36 x+36 y^{2}-18 y+11=0} \\ {36 x^{2}+36 y^{2}-36 x-18 y+11=0}\end{array}\) 

Ans : 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) = (1, 1) and radius (r) = \(\sqrt{2}\)
Therefore, the equation of the circle is
\(\begin{array}{l}{(x-1)^{2}+(y-1)^{2}=(\sqrt{2})^{2}} \\ {x^{2}-2 x+1+y^{2}-2 y+1=2} \\ {x^{2}+y^{2}-2 x-2 y=0}\end{array}\) 

Ans : 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
\(\begin{array}{l}{(x+a)^{2}+(y+b)^{2}=\left(\sqrt{a^{2}-b^{2}}\right)^{2}} \\ {x^{2}+2 a x+a^{2}+y^{2}+2 b y+b^{2}=a^{2}-b^{2}} \\ {x^{2}+y^{2}+2 a x+2 b y+2 b^{2}=0}\end{array}\)