NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.1) Exercise 11.1

Q.1 In each of the following Exercises 1 to 5, find the equation of the circle with
1. centre (0,2) and radius 2
2. centre (–2,3) and radius 4
3.

$\mathbf{centre}\mathbf{(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{)}\mathbf{and}\mathbf{}\mathbf{radius}\mathbf{}\frac{\mathbf{1}}{\mathbf{12}}$

4.

$\mathbf{centre}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{and}\mathbf{}\mathbf{radius}\mathbf{}\sqrt{\mathbf{2}}$

5.

$\mathbf{centre}\mathbf{(}\mathbf{-}\mathbf{a}\mathbf{,}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{}\mathbf{and}\mathbf{}\mathbf{radius}\mathbf{}\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}}$

Ans.
1.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\\ \mathrm{Equation}\text{of circle with centre}(0,2)\text{and radius 2 is:}\\ {(\mathrm{x}-0)}^2+{(\mathrm{y}-2)}^2=2^2\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{y}+4=4\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{y}=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

2.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\\ \mathrm{Equation}\text{of circle with centre}(-2,3)\text{and radius 4 is:}\\ {(\mathrm{x}+2)}^2+{(\mathrm{y}-3)}^2=4^2\\ \Rightarrow {\mathrm{x}}^2+4\mathrm{x}+4+{\mathrm{y}}^2-6\mathrm{y}+9=16\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-6\mathrm{y}-3=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

3.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\\ \mathrm{Equation}\text{of circle with centre}(\frac{1}{2},\frac{1}{4})\text{and radius}\frac{1}{12}\text{is:}\\ {(\mathrm{x}-\frac{1}{2})}^2+{(\mathrm{y}-\frac{1}{4})}^2={\left(\frac{1}{12}\right)}^2\\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}+\frac{1}{4}+{\mathrm{y}}^2-\frac{1}{2}\mathrm{y}+\frac{1}{16}=\frac{1}{144}\\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}+{\mathrm{y}}^2-\frac{1}{2}\mathrm{y}+\frac{1}{4}+\frac{1}{16}-\frac{1}{144}=0\\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}+{\mathrm{y}}^2-\frac{1}{2}\mathrm{y}+\frac{11}{36}=0\\ \Rightarrow 36{\mathrm{x}}^2+36{\mathrm{y}}^2-36\mathrm{x}-18\mathrm{y}+11=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

4.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\\ \mathrm{Equation}\text{of circle with centre}(1,1)\text{and radius}\sqrt{\text{2}}\text{is:}\end{array}$ \begin{array}{}\end{array} $\begin{array}{l}\end{array}$

$\begin{array}{l}{(\mathrm{x}-1)}^2+{(\mathrm{y}-1)}^2={\left(\sqrt{\text{2}}\right)}^2\\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}+1+{\mathrm{y}}^2-2\mathrm{y}+1=2\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{y}=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

5.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\\ \mathrm{Equation}\text{of circle with centre}(-\mathrm{a},-\mathrm{b})\text{and radius}\sqrt{{\text{a}}^{\text{2}}-{\mathrm{b}}^2}\text{is:}\\ {(\mathrm{x}+\mathrm{a})}^2+{(\mathrm{y}+\mathrm{b})}^2={\left(\sqrt{{\text{a}}^{\text{2}}-{\mathrm{b}}^2}\right)}^2\\ {\mathrm{x}}^2+2\mathrm{ax}+{\mathrm{a}}^2+{\mathrm{y}}^2+2\mathrm{by}+{\mathrm{b}}^2={\text{a}}^{\text{2}}-{\mathrm{b}}^2\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{ax}+2\mathrm{by}+2{\mathrm{b}}^2=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

Q.2 In each of the following Exercises 6 to 9, find the centre and radius of the circles.

6. (x + 5)² + (y – 3)² = 36
7. x² + y² – 4x – 8y – 45 = 0
8. x² + y² – 8x + 10y – 12 = 0
9. 2x² + 2y² – x = 0

Ans.

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ {\text{(x+5)}}^{\text{2}}\text{+(y}-{\text{3)}}^{\text{2}}\text{=36}...\left(\text{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\text{and we get}-\\ \mathrm{h}=-5,\mathrm{k}=3{\text{and r}}^{\text{2}}=\text{36}\end{array}$

$\begin{array}{l}\Rightarrow \mathrm{h}=-5,\mathrm{k}=3\text{and r}=\text{6}\\ \text{Therefore, centre is}(-5,3)\text{and radius is 6.}\end{array}$

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-8\mathrm{y}-45=0\\ \Rightarrow ({\mathrm{x}}^2-4\mathrm{x})+({\mathrm{y}}^2-8\mathrm{y})-45=0\\ \Rightarrow ({\mathrm{x}}^2-4\mathrm{x}+2^2-2^2)+({\mathrm{y}}^2-8\mathrm{y}+4^2-4^2)-45=0\\ \Rightarrow ({\mathrm{x}}^2-4\mathrm{x}+2^2)-4+({\mathrm{y}}^2-8\mathrm{y}+4^2)-16-45=0\\ \Rightarrow {({\mathrm{x}}^2-2)}^2+{({\mathrm{y}}^2-4)}^2-65=0\\ \Rightarrow {({\mathrm{x}}^2-2)}^2+{({\mathrm{y}}^2-4)}^2={\left(\sqrt{65}\right)}^2...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\text{and we get}-\\ \mathrm{h}=2,\mathrm{k}=4\text{and r}=\sqrt{65}\\ \text{Therefore, centre is}(2,4)\text{and radius is}\sqrt{\text{65}}\text{.}\end{array}$

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ {\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}+10\mathrm{y}-12=0\\ \Rightarrow ({\mathrm{x}}^2-8\mathrm{x})+({\mathrm{y}}^2+10\mathrm{y})-12=0\\ \Rightarrow ({\mathrm{x}}^2-8\mathrm{x}+4^2-4^2)+({\mathrm{y}}^2+10\mathrm{y}+5^2-5^2)-12=0\\ \Rightarrow ({\mathrm{x}}^2-8\mathrm{x}+4^2)-16+({\mathrm{y}}^2+10\mathrm{y}+5^2)-25-12=0\end{array}$

$\begin{array}{l}\Rightarrow {({\mathrm{x}}^2-4)}^2+{({\mathrm{y}}^2+5)}^2-53=0\\ \Rightarrow {({\mathrm{x}}^2-4)}^2+{({\mathrm{y}}^2+5)}^2={\left(\sqrt{53}\right)}^2...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\text{and we get}-\\ \mathrm{h}=4,\mathrm{k}=-5\text{and r}=\sqrt{53}\\ \text{Therefore, centre is}(4,-5)\text{and radius is}\sqrt{\text{53}}\text{.}\end{array}$

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}2x}}^{\text{2}}+{\text{2y}}^{\text{2}}-\text{x}=0\\ \Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}x}}^{\text{2}}+{\text{y}}^{\text{2}}-\frac{1}{\text{2}}\text{x}=0\\ \Rightarrow ({\mathrm{x}}^2-\frac{1}{2}\mathrm{x})+{\mathrm{y}}^2=0\\ \Rightarrow ({\mathrm{x}}^2-\frac{1}{2}\mathrm{x}+\frac{1}{4^2}-\frac{1}{4^2})+{\mathrm{y}}^2=0\\ \Rightarrow ({\mathrm{x}}^2-\frac{1}{2}\mathrm{x}+\frac{1}{4^2})+{\mathrm{y}}^2=\frac{1}{2^2}\\ \Rightarrow {(\mathrm{x}-\frac{1}{4})}^2+{\mathrm{y}}^2=\frac{1}{4^2}...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\text{and we get}-\\ \mathrm{h}=\frac{1}{4},\mathrm{k}=0\text{and r}=\frac{1}{4}\end{array}$

$\text{Therefore, centre is}(\frac{1}{4},0)\text{and radius is}\frac{1}{4}\text{.}$

Q.3 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.

Ans.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{i}\right)\\ \mathrm{Circle}\left(\mathrm{i}\right)\text{is passing through}(4,1)\text{and}(6,5)\text{then,}\\ {(4-\mathrm{h})}^2+{(1-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{ii}\right)\\ \mathrm{and}{(6-\mathrm{h})}^2+{(5-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{\hspace{0.17em}and equation}\left(\mathrm{iii}\right),\text{we get}\\ {(4-\mathrm{h})}^2+{(1-\mathrm{k})}^2={(6-\mathrm{h})}^2+{(5-\mathrm{k})}^2\\ \Rightarrow 16-8\mathrm{h}+{\mathrm{h}}^2+1-2\mathrm{k}+{\mathrm{k}}^2=36-12\mathrm{h}+{\mathrm{h}}^2+25-10\mathrm{k}+{\mathrm{k}}^2\\ \Rightarrow 4\mathrm{h}+8\mathrm{k}=44\\ \Rightarrow \mathrm{h}+2\mathrm{k}=11...\left(\mathrm{iv}\right)\\ \mathrm{Since},\text{ centre}(\mathrm{h},\mathrm{k})\text{lies on the line 4x}+\mathrm{y}=16.\\ \mathrm{So},4\mathrm{h}+\mathrm{k}=16...\left(\mathrm{v}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{iv}\right)\text{and equation}\left(\mathrm{v}\right),\text{we get}\\ \text{h}=\text{3 and k}=\text{4}\\ \text{Putting value of h and k in equation}\left(\mathrm{ii}\right),\text{we get}\\ {(4-3)}^2+{(1-4)}^2={\mathrm{r}}^2\\ \Rightarrow {\mathrm{r}}^2=1+9\\ =10\end{array}$

$\begin{array}{l}\mathrm{Then},\text{equation of circle is:}\\ {(\mathrm{x}-3)}^2+{(\mathrm{y}-4)}^2=10\\ \Rightarrow {\mathrm{x}}^2-6\mathrm{x}+9+{\mathrm{y}}^2-8\mathrm{y}+16=10\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-8\mathrm{y}+15=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

Q.4 Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0.

Ans.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{i}\right)\\ \mathrm{Circle}\left(\mathrm{i}\right)\text{is passing through}(2,3)\text{and}(-1,1)\text{then,}\\ {(2-\mathrm{h})}^2+{(3-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{ii}\right)\\ \mathrm{and}{(-1-\mathrm{h})}^2+{(1-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{\hspace{0.17em}and equation}\left(\mathrm{iii}\right),\text{we get}\\ {(2-\mathrm{h})}^2+{(3-\mathrm{k})}^2={(-1-\mathrm{h})}^2+{(1-\mathrm{k})}^2\\ \Rightarrow 4-4\mathrm{h}+{\mathrm{h}}^2+9-6\mathrm{k}+{\mathrm{k}}^2=1+2\mathrm{h}+{\mathrm{h}}^2+1-2\mathrm{k}+{\mathrm{k}}^2\\ \Rightarrow -6\mathrm{h}-4\mathrm{k}=-11\\ \Rightarrow 6\mathrm{h}+4\mathrm{k}=11...\left(\mathrm{iv}\right)\\ \mathrm{Since},\text{ centre}(\mathrm{h},\mathrm{k})\text{lies on the line x}-3\mathrm{y}=11.\\ \mathrm{So},\mathrm{h}-3\mathrm{k}=11...\left(\mathrm{v}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{iv}\right)\text{and equation}\left(\mathrm{v}\right),\text{we get}\\ \text{h}=\frac{7}{2}\text{and k}=-\frac{\text{5}}{2}\end{array}$

$\begin{array}{l}\text{Putting value of h and k in equation}\left(\mathrm{ii}\right),\text{we get}\\ {(2-\frac{7}{2})}^2+{(3+\frac{\text{5}}{2})}^2={\mathrm{r}}^2\\ \Rightarrow {\mathrm{r}}^2=\frac{9}{4}+\frac{121}{4}\\ =\frac{130}{4}\\ \mathrm{Then},\text{equation of circle is:}\\ {(\mathrm{x}-\frac{7}{2})}^2+{(\mathrm{y}-\frac{\text{5}}{2})}^2=\frac{130}{4}\\ \Rightarrow {\mathrm{x}}^2-7\mathrm{x}+\frac{49}{4}+{\mathrm{y}}^2+5\mathrm{y}+\frac{25}{4}=\frac{130}{4}\\ \Rightarrow 4{\mathrm{x}}^2+4{\mathrm{y}}^2-28\mathrm{x}+20\mathrm{y}=56\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-7\mathrm{x}+5\mathrm{y}-14=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

Q.5 Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).

Ans.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{a},0)\text{and radius 5 is:}\\ {(\mathrm{x}-\mathrm{a})}^2+{(\mathrm{y}-0)}^2=5^2\\ \Rightarrow {(\mathrm{x}-\mathrm{a})}^2+{\mathrm{y}}^2=5^2...\left(\mathrm{i}\right)\\ \mathrm{Since},\text{circle}\left(\mathrm{i}\right)\text{passes through the point}(2,3).\mathrm{So},\\ {(2-\mathrm{a})}^2+3^2=5^2\\ \Rightarrow 4-4\mathrm{a}+{\mathrm{a}}^2+9=25\\ \Rightarrow {\mathrm{a}}^2-4\mathrm{a}=25-13\end{array}$

$\begin{array}{l}\Rightarrow {\mathrm{a}}^2-4\mathrm{a}-12=0\\ \Rightarrow (\mathrm{a}-6)(\mathrm{a}+2)=0\\ \Rightarrow \mathrm{a}=6,-2\\ \mathrm{Case}1:\mathrm{When}\text{centre of circle is}(6,0):\\ \mathrm{The}\text{equation of circle is:-}\\ {(\mathrm{x}-6)}^2+{(\mathrm{y}-0)}^2=5^2\\ \Rightarrow {\mathrm{x}}^2-12\mathrm{x}+36+{\mathrm{y}}^2=25\\ \Rightarrow {\mathrm{x}}^2-12\mathrm{x}+{\mathrm{y}}^2+11=0\\ \mathrm{Case}2:\mathrm{When}\text{centre of circle is}(-2,0):\\ \mathrm{The}\text{equation of circle is:-}\\ {(\mathrm{x}+2)}^2+{(\mathrm{y}-0)}^2=5^2\\ \Rightarrow {\mathrm{x}}^2+4\mathrm{x}+4+{\mathrm{y}}^2=25\\ \Rightarrow {\mathrm{x}}^2+4\mathrm{x}+{\mathrm{y}}^2-21=0\\ \mathrm{Thus},\text{the equation of circle is}\\ {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-21=0\text{ or}{\mathrm{x}}^2+{\mathrm{y}}^2-12\mathrm{x}+11=0.\end{array}$

Q.6 Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

Ans.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{i}\right)\\ \mathrm{Circle}\text{passes through the points}(0,0),(\mathrm{a},0)\text{and}(0,\mathrm{b}),\text{so}\\ {(0-\mathrm{h})}^2+{(0-\mathrm{k})}^2={\mathrm{r}}^2\\ \Rightarrow {\mathrm{h}}^2+{\mathrm{k}}^2={\mathrm{r}}^2...\left(\mathrm{ii}\right)\end{array}$ $\begin{array}{l}{}^{}\end{array}$

$\begin{array}{l}{(\mathrm{a}-\mathrm{h})}^2+{(0-\mathrm{k})}^2={\mathrm{r}}^2\\ \Rightarrow {\mathrm{a}}^2-2\mathrm{ah}+{\mathrm{h}}^2+{\mathrm{k}}^2={\mathrm{r}}^2...\left(\mathrm{iii}\right)\\ {(0-\mathrm{h})}^2+{(\mathrm{b}-\mathrm{k})}^2={\mathrm{r}}^2\\ \Rightarrow {\mathrm{h}}^2+{\mathrm{b}}^2-2\mathrm{bk}+{\mathrm{k}}^2={\mathrm{r}}^2...\left(\mathrm{iv}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we get}\\ {\mathrm{h}}^2+{\mathrm{k}}^2={\mathrm{a}}^2-2\mathrm{ah}+{\mathrm{h}}^2+{\mathrm{k}}^2\\ \Rightarrow {\mathrm{a}}^2-2\mathrm{ah}=0\\ \Rightarrow \mathrm{a}(\mathrm{a}-2\mathrm{h})=0\\ \Rightarrow \mathrm{a}-2\mathrm{h}=0,\mathrm{a}\ne 0\\ \Rightarrow \mathrm{h}=\frac{\mathrm{a}}{2}\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iv}\right),\text{we get}\\ {\mathrm{h}}^2+{\mathrm{k}}^2={\mathrm{h}}^2+{\mathrm{b}}^2-2\mathrm{bk}+{\mathrm{k}}^2\\ \Rightarrow {\mathrm{b}}^2-2\mathrm{bk}=0\\ \Rightarrow \mathrm{b}(\mathrm{b}-2\mathrm{k})=0\\ \Rightarrow \mathrm{b}-2\mathrm{k}=0,\mathrm{b}\ne 0\\ \Rightarrow \mathrm{k}=\frac{\mathrm{b}}{2}\\ \mathrm{Putting}\text{ values of h and k in equation}\left(\mathrm{ii}\right),\text{we get}\\ {\left(\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\mathrm{b}}{2}\right)}^2={\mathrm{r}}^2\\ \frac{{\mathrm{a}}^2}{4}+\frac{{\mathrm{b}}^2}{4}={\mathrm{r}}^2\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right):\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\end{array}$

$\begin{array}{l}{(\mathrm{x}-\frac{\mathrm{a}}{2})}^2+{(\mathrm{y}-\frac{\mathrm{b}}{2})}^2=\frac{{\mathrm{a}}^2}{4}+\frac{{\mathrm{b}}^2}{4}\\ \Rightarrow {\mathrm{x}}^2-\mathrm{ax}+\frac{{\mathrm{a}}^2}{4}+{\mathrm{y}}^2-\mathrm{by}+\frac{{\mathrm{b}}^2}{4}=\frac{{\mathrm{a}}^2}{4}+\frac{{\mathrm{b}}^2}{4}\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{ax}-\mathrm{by}=0\\ \mathrm{This}\text{is the required equation of the circle.}\end{array}$

Q.7 Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Ans.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(2,2)\text{and radius r is:}\\ {(\mathrm{x}-2)}^2+{(\mathrm{y}-2)}^2={\mathrm{r}}^2...\left(\mathrm{i}\right)\\ \mathrm{Circle}\text{passes through the point}(4,5),\text{so}\\ {(4-2)}^2+{(5-2)}^2={\mathrm{r}}^2\\ \Rightarrow 4+9={\mathrm{r}}^2\\ \Rightarrow 13={\mathrm{r}}^2\\ \mathrm{Putting}{\text{value of r}}^{\text{2}}\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ {(\mathrm{x}-2)}^2+{(\mathrm{y}-2)}^2=13\\ \Rightarrow {\mathrm{x}}^2-4\mathrm{x}+4+{\mathrm{y}}^2-4\mathrm{y}+4=13\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-4\mathrm{y}=5\\ \mathrm{This}\text{is the equation of required circle.}\end{array}$

Q.8 Does the point (–2.5, 3.5) lie inside, outside or on the circle x² + y² = 25?

Ans.

$\begin{array}{l}\mathrm{The}\text{equation of given circle is:}\\ {\text{x}}^{\text{2}}+{\text{y}}^{\text{2}}=\text{25}\end{array}$

$\begin{array}{l}\text{Putting x}=\text{2}.\text{5},\text{y}=\text{3}.\text{5 in L.H.S. of given equation, we get}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{2}}+{\text{y}}^{\text{2}}={(\text{2}.\text{5})}^{\text{2}}+{(\text{3}.\text{5})}^{\text{2}}\\ =6.25+12.25\\ =18.50<25=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{point}(–\text{2}.\text{5},\text{3}.\text{5})\text{lies inside of the circle.}\\ \text{As distance of the point from centre is less than radius}\\ \text{of the circle.}\end{array}$