NCERT Solutions for Class 11 Maths Chapter 11: Conic Sections - Exercise 11.1

Exercise 11.1

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

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 $(\mathrm{h}, \mathrm{k})=(0,2)$ and radius $\mathrm{r}=2$.

Therefore, equation of the circle is

$\begin{aligned} &(x-0)^{2}+(y-2)^{2}=2^{2} \\ &x^{2}+y^{2}+4-4 y=4 \\ &x^{2}+y^{2}-4 y=0 \end{aligned}$

2. Find the equation of the circle with centre $(-2,3)$ and radius 4 .

Ans: The equation of a circle with centre $(h, k)$ and radius $r$ is given as

$(\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{r}^{2} $

It is given that centre $(\mathrm{h}, \mathrm{k})=(-2,3)$ and radius $\mathrm{r}=4$.

Therefore, equation of the circle is

$\begin{aligned} &(x+2)^{2}+(y-3)^{2}=4^{2} \\ &x^{2}+4 x+4+y^{2}+9-6 y=16 \\ &x^{2}+y^{2}+4 x-6 y-3=0 \end{aligned} $

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

Ans: The equation of a circle with centre $(h, k)$ and radius $r$ is given as

$ (\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{r}^{2}$

It is given that centre $(h, k)=\left(\frac{1}{2}, \frac{1}{4}\right)$ and radius $r=\frac{1}{12}$.

Therefore, equation of the circle is 

$\begin{aligned} &\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{4}\right)^{2}=\left(\frac{1}{12}\right)^{2} \\ &x^{2}-x+\frac{1}{4}+y^{2}+\frac{1}{16}-\frac{y}{2}=\frac{1}{144} \\ &144 x^{2}-144 x+36+144 y^{2}-72 y+9-1=0 \\ &144 x^{2}-144 x+144 y^{2}-72 y+44=0 \\ &36 x^{2}+36 y^{2}-36 x-18 y+11=0 \end{aligned}$

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

Ans: The equation of a circle with centre $(\mathrm{h}, \mathrm{k})$ and radius $\mathrm{r}$ is given as

$(\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{r}^{2} $

It is given that centre $(h, k)=(1,1)$ and radius $r=\sqrt{2}$.

Therefore, equation of the circle is

$\begin{aligned} &(x-1)^{2}+(y-1)^{2}=(\sqrt{2})^{2} \\ &x^{2}-2 x+1+y^{2}-2 y+1=2 \\ &x^{2}+y^{2}-2 x-2 y=0 \end{aligned}$

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

Ans: The equation of a circle with centre $(h, k)$ and radius $r$ is given as

$(\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{r}^{2}$

It is given that centre $(h, k)=(-a,-b)$ and radius $r=\sqrt{a^{2}-b^{2}}$.

Therefore, equation of the circle is

$\begin{aligned} &(x+a)^{2}+(y+b)^{2}=\left(\sqrt{a^{2}-b^{2}}\right)^{2} \\ &x^{2}+2 a x+a^{2}+y^{2}+2 b y+b^{2}=a^{2}-b^{2} \\ &x^{2}+y^{2}+2 a x+2 b y+2 b^{2}=0 \end{aligned}$

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

Ans: The equation of the given circle is $(x+5)^{2}+(y-3)^{2}=36$.

The equation can be written as $[x-(-5)]^{2}+[y-(3)]^{2}=(6)^{2} \ldots \ldots$. (Equation 1) Let us compare it with the general equation of circle i.e.,

$(x-h)^{2}+(y-k)^{2}=r^{2} \ldots \ldots .($ Equation 2$)$

On observing (Equation 1) and (Equation 2), we get

$\mathrm{h}=-5, \mathrm{k}=3$ and $\mathrm{r}=6$

Thus, the coordinates of the centre is $(-5,6)$ and radius is 6 .

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

Ans: The equation of the given circle is $x^{2}+y^{2}-4 x-8 y-45=0$.

The equation can be written as

$\begin{aligned} &\left(x^{2}-4 x\right)+\left(y^{2}-8 y\right)=45 \\ &\Rightarrow\left[x^{2}-4 x+(2)^{2}\right]+\left[y^{2}-8 y+(4)^{2}\right]-(2)^{2}-(4)^{2}=45 \\ &\Rightarrow(x-2)^{2}+(y-4)^{2}=65 \\ &\Rightarrow(x-2)^{2}+(y-4)^{2}=(\sqrt{65})^{2} \ldots \ldots \text {.(Equation 1) } \end{aligned}$

Let us compare it with the general equation of circle i.e.,

$(x-h)^{2}+(y-k)^{2}=r^{2} \ldots \ldots .(\text { Equation } 2)$

On observing (Equation 1) and (Equation 2), we get

$\mathrm{h}=2, \mathrm{k}=4 \text { and } \mathrm{r}=\sqrt{65}$

Thus, the coordinates of the centre are $(2,4)$ and radius is $\sqrt{65}$.

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

Ans: The equation of the given circle is $x^{2}+y^{2}-8 x-10 y-12=0$.

The equation can be written as

$\begin{aligned} &\left(x^{2}-8 x\right)+\left(y^{2}-10 y\right)=12 \\ &\Rightarrow\left[x^{2}-8 x+(4)^{2}\right]+\left[y^{2}-10 y+(5)^{2}\right]-(4)^{2}-(5)^{2}=12 \\ &\Rightarrow(x-4)^{2}+(y-5)^{2}=53 \\ &\Rightarrow(x-4)^{2}+(y-5)^{2}=(\sqrt{53})^{2} \ldots \ldots .(\text { Equation 1) } \end{aligned}$

Let us compare it with the general equation of circle i.e., $(x-h)^{2}+(y-k)^{2}=r^{2} \ldots \ldots .($ Equation 2$)$ On observing (Equation 1) and (Equation 2), we get $\mathrm{h}=4, \mathrm{k}=5$ and $\mathrm{r}=\sqrt{53}$

Thus, the coordinates of the centre are $(4,5)$ and radius is $\sqrt{53}$.

9. Find the centre and radius of the circle $2 x^{2}+2 y^{2}-x=0$

Ans: The equation of the given circle is $2 \mathrm{x}^{2}+2 \mathrm{y}^{2}-\mathrm{x}=0$.

The equation can be written as

$\begin{aligned} &2\left(x^{2}-\frac{x}{2}\right)+2\left(y^{2}\right)=0 \\ &\Rightarrow\left[x^{2}-\frac{x}{2}+\left(\frac{1}{4}\right)^{2}\right]+\left[y^{2}\right]-\left(\frac{1}{4}\right)^{2}=0 \\ &\Rightarrow\left(x-\frac{1}{4}\right)^{2}+(y)^{2}=\frac{1}{16} \\ &\Rightarrow\left(x-\frac{1}{4}\right)^{2}+(y-0)^{2}=\left(\frac{1}{4}\right)^{2} \ldots \ldots .(\text { Equation 1) } \end{aligned}$

Let us compare it with the general equation of circle i.e.,

$(x-h)^{2}+(y-k)^{2}=r^{2} \ldots \ldots .(\text { Equation } 2)$

On observing (Equation 1) and (Equation 2), we get

$\mathrm{h}=\frac{1}{4}, \mathrm{k}=0 \text { and } \mathrm{r}=\frac{1}{4}$

Thus, the coordinates of the centre are $\left(\frac{1}{4}, 0\right)$ and radius is $\frac{1}{4}$.

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

Ans: Let the equation of the required circle be $(x-h)^{2}+(y-k)^{2}=r^{2}$.

Since the circle passes through point $(4,1)$ and $(6,5)$,

$\begin{aligned} &(4-\mathrm{h})^{2}+(1-\mathrm{k})^{2}=\mathrm{r}^{2} \\ &\Rightarrow 16+\mathrm{h}^{2}-8 \mathrm{~h}+1+\mathrm{k}^{2}-2 \mathrm{k}=\mathrm{r}^{2} \\ &\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}-8 \mathrm{~h}-2 \mathrm{k}+17=\mathrm{r}^{2} \ldots \end{aligned}$ $\begin{aligned} &(6-\mathrm{h})^{2}+(5-\mathrm{k})^{2}=\mathrm{r}^{2} \\ &\Rightarrow 36+\mathrm{h}^{2}-12 \mathrm{~h}+25+\mathrm{k}^{2}-10 \mathrm{k}=\mathrm{r}^{2} \\ &\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}-12 \mathrm{~h}-10 \mathrm{k}+61=\mathrm{r}^{2} \ldots(2) \end{aligned}$

Since the centre $(h, k)$ of the circle lies on line $4 x+y=16$,

$4 \mathrm{~h}+\mathrm{k}=16 \ldots \text { (3) }$

From equations (1) and (2), we obtain

$\begin{aligned} &\mathrm{h}^{2}+\mathrm{k}^{2}-8 \mathrm{~h}-2 \mathrm{k}+17 \\ &\Rightarrow 4 \mathrm{~h}+8 \mathrm{k}=44 \\ &\Rightarrow \mathrm{h}+2 \mathrm{k}=11 \ldots \text { (4) } \end{aligned}$ $\begin{aligned} & \Rightarrow 4 \mathrm{~h}+8 \mathrm{k}=44 \end{aligned} $

On solving equations (3) and (4), we obtain $\mathrm{h}=3$ and $\mathrm{k}=4$.

On substituting the values of $h$ and $k$ in equation (1), we obtain

$\begin{aligned} &(4-3)^{2}+(1-4)^{2}=r^{2} \\ &\Rightarrow(1)^{2}+(-3)^{2}=r^{2} \\ &\Rightarrow 1+9=r^{2} \\ &\Rightarrow r^{2}=10 \end{aligned}$

Thus, the equation of the required circle is

$\begin{aligned} &(x-3)^{2}+(y-4)^{2}=10 \\ &x^{2}-6 x+9+y^{2}-8 y+16=10 \\ &\Rightarrow x^{2}+y^{2}-6 x-8 y+15=0 \end{aligned}$

11. Find the equation of the circle passing through the points $(2,3)$ and} $(-1,1)$ and whose centre is on the line $\mathbf{x}-3 \mathbf{y}-11=\mathbf{0}$.

Ans: Let the equation of the required circle be $(x-h)^{2}+(y-k)^{2}=r^{2}$.

Since the circle passes through point $(2,3)$ and $(-1,1)$,

$\begin{aligned} &(2-\mathrm{h})^{2}+(3-\mathrm{k})^{2}=\mathrm{r}^{2} \\ &\Rightarrow 4+\mathrm{h}^{2}-4 \mathrm{~h}+9+\mathrm{k}^{2}-6 \mathrm{k}=\mathrm{r}^{2} \\ &\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}-4 \mathrm{~h}-6 \mathrm{k}+13=\mathrm{r}^{2} \ldots \\ &(-1-\mathrm{h})^{2}+(1-\mathrm{k})^{2}=\mathrm{r}^{2} \\ &\Rightarrow 1+\mathrm{h}^{2}+2 \mathrm{~h}+1+\mathrm{k}^{2}-2 \mathrm{k}=\mathrm{r}^{2} \\ &\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}+2 \mathrm{~h}-2 \mathrm{k}+2=\mathrm{r}^{2} \ldots \end{aligned}$

Since the centre $(h, k)$ of the circle lies on line $x-3 y-11=0$, $h-3 k-11=0 \ldots$ (3)

From equations (1) and (2), we obtain

$h^{2}+k^{2}-4 h-6 k+13=h^{2}+k^{2}+2 h-2 k+2$

$\Rightarrow-6 \mathrm{~h}-4 \mathrm{k}=-11$

$\Rightarrow 6 h+4 k=11 \ldots$ (4)

On solving equations (3) and (4), we obtain $\mathrm{h}=\frac{7}{2}$ and $\mathrm{k}=\frac{-5}{2}$.

On substituting the values of $h$ and $k$ in equation (1), we obtain

$\begin{aligned} &\left(2-\frac{7}{2}\right)^{2}+\left(3+\frac{5}{2}\right)^{2}=r^{2} \\ &\Rightarrow\left(\frac{4-7}{2}\right)^{2}+\left(\frac{6+5}{2}\right)^{2}=r^{2} \\ &\Rightarrow\left(\frac{-3}{2}\right)^{2}+\left(\frac{11}{2}\right)^{2}=r^{2} \\ &\Rightarrow \frac{9}{4}+\frac{121}{4}=r^{2} \\ &\Rightarrow \frac{130}{4}=r^{2} \end{aligned}$

Thus, the equation of the required circle is

$\begin{aligned} &\left(x-\frac{7}{2}\right)^{2}+\left(y+\frac{5}{2}\right)^{2}=\frac{65}{2} \\ &\Rightarrow x^{2}+\frac{49}{4}-7 x+y^{2}+\frac{25}{4}+5 y=\frac{65}{2} \\ &\Rightarrow 4 x^{2}+49-28 x+4 y^{2}+25+20 y=130 \\ &\Rightarrow 4 x^{2}+4 y^{2}-28 x+20 y-56=0 \\ &\Rightarrow 4\left(x^{2}+y^{2}-7 x+5 y-14\right)=0 \\ &\Rightarrow x^{2}+y^{2}-7 x+5 y-14=0 \end{aligned}$

12. Find the equation of circle with radius 5 whose centre lies on $\mathrm{x}$-axis and passes through the point $(2,3)$.

Ans: Let the equation of required circle be $(x-h)^{2}+(y-k)^{2}=r^{2}$

Since the centre lies on the $x$-axis, $k=0$.

Also, we are given that $r=0$. Hence, we get equation as $(\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-0)^{2}=5^{2} \ldots(1)$

Also, it is given that the circle passes through the point $(2,3)$,

This means that the point satisfies equation (1).

Substituting point $(2,3)$ in equation $(1)$, we get,

$(2-\mathrm{h})^{2}+(3-0)^{2}=5^{2}$

$\Rightarrow 4-4 \mathrm{~h}+\mathrm{h}^{2}+9=25$

$\Rightarrow \mathrm{h}^{2}-4 \mathrm{~h}-12=0$

$\Rightarrow \mathrm{h}^{2}-6 \mathrm{~h}+2 \mathrm{~h}-12=0$

$\Rightarrow \mathrm{h}(\mathrm{h}-6)+2(\mathrm{~h}-6)=0$

$\Rightarrow(\mathrm{h}+2)(\mathrm{h}-6)=0$

$\Rightarrow \mathrm{h}=-2$ or $\mathrm{h}=6$

When $\mathrm{h}=-2$, the equation of circle becomes,

$(x+2)^{2}+(y-0)^{2}=5^{2}$

$\Rightarrow \mathrm{x}^{2}+4 \mathrm{x}+4+\mathrm{y}^{2}=25$

$\Rightarrow \mathrm{x}^{2}+4 \mathrm{x}+\mathrm{y}^{2}-21=0$

When $\mathrm{h}=6$, the equation of circle becomes,

$(x-6)^{2}+(y-0)^{2}=5^{2}$

$\Rightarrow x^{2}-12 x+36+y^{2}=25$

$\Rightarrow x^{2}-12 x+y^{2}+11=0$

13. Find the equation of circle passing through $(0,0)$ and making intercepts a and bon the coordinate axes.

Ans: Let the equation of required circle be $(x-h)^{2}+(y-k)^{2}=r^{2} \ldots(1)$

We are given that the circle passes through the point $(0,0)$,

This means the point satisfies equation (1),

$(x-h)^{2}+(y-k)^{2}=r^{2}$

$\Rightarrow(0-\mathrm{h})^{2}+(0-\mathrm{k})^{2}=\mathrm{r}^{2}$

$\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{r}^{2} \ldots(2)$

From Equation (1) and (2), we get $(x-h)^{2}+(y-k)^{2}=h^{2}+k^{2}$

It is given that the circle makes $x$-intercept at $(a, 0)$ and y-intercept at $(0, b)$. Substituting the points in Equation (3),

$\begin{aligned} &(\mathrm{a}-\mathrm{h})^{2}+(0-\mathrm{k})^{2}=\mathrm{h}^{2}+\mathrm{k}^{2} \\ &\Rightarrow \mathrm{a}^{2}-2 \mathrm{ah}+\mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{h}^{2}+\mathrm{k}^{2} \\ &\Rightarrow \mathrm{a}^{2}-2 \mathrm{ah}=0 \\ &\Rightarrow \mathrm{a}(\mathrm{a}-2 \mathrm{~h})=0 \\ &\Rightarrow \mathrm{a}=0 \text { or } \mathrm{a}=2 \mathrm{~h} \end{aligned}$

However, $a \neq 0 ;$ Hence, $h=\frac{a}{2}$.

Similarly,

$\begin{aligned} &(0-\mathrm{h})^{2}+(\mathrm{b}-\mathrm{k})^{2}=\mathrm{h}^{2}+\mathrm{k}^{2} \\ &\Rightarrow \mathrm{h}^{2}+\mathrm{b}^{2}-2 \mathrm{bk}+\mathrm{k}^{2}=\mathrm{h}^{2}+\mathrm{k}^{2} \\ &\Rightarrow \mathrm{b}^{2}-2 \mathrm{bk}=0 \\ &\Rightarrow \mathrm{b}(\mathrm{b}-2 \mathrm{k})=0 \\ &\Rightarrow \mathrm{b}=0 \text { or } \mathrm{b}=2 \mathrm{k} \end{aligned}$

However, $b \neq 0$; Hence, $k=\frac{b}{2}$.

Hence, the equation of required circle is

$\begin{aligned} &\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{b}{2}\right)^{2}=\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2} \\ &\Rightarrow x^{2}-a x+\frac{a^{2}}{4}+y^{2}-b y+\frac{b^{2}}{4}=\frac{a^{2}}{4}+\frac{b^{2}}{4} \\ &\Rightarrow x^{2}-a x+y^{2}-b y=0 \\ &\Rightarrow x^{2}+y^{2}-a x-b y=0 \end{aligned}$

14. Find the equation of circle with centre $(2,2)$ and passes through the point $(4,5)$.

Ans: Let the equation of required circle be $(x-h)^{2}+(y-k)^{2}=r^{2} \ldots(1)$

The centre of the circle is given as $(h, k)=(2,2)$ Since the circle passes through the point $(4,5)$, the radius $r$ of the circle will be given by the distance between centre and the point.

$\begin{aligned} &\mathrm{r}=\sqrt{(4-2)^{2}+(5} \\ &\Rightarrow \mathrm{r}=\sqrt{2^{2}+3^{3}} \\ &\Rightarrow \mathrm{r}=\sqrt{4+9} \\ &\Rightarrow \mathrm{r}=\sqrt{13} \end{aligned}$ Hence, the equation of the circle is $\begin{aligned} &(x-h)^{2}+(y-k)^{2}=r^{2} \\ &\Rightarrow(x-2)^{2}+(y-2)^{2}=(\sqrt{13})^{2} \\ &\Rightarrow x^{2}-4 x+4+y^{2}-4 y+4=13 \\ &\Rightarrow x^{2}+y^{2}-4 x-4 y-5=0 \end{aligned}$

15. Does the point $(-2.5,3.5)$ lie inside, outside or on the circle $x^{2}+y^{2}=25$ ?

Ans: The equation of the circle is $x^{2}+y^{2}=25$, which can be written as $(x-0)^{2}+(y-0)^{2}=5^{2} .$

This equation is of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$.

Hence, centre $(\mathrm{h}, \mathrm{k})=(0,0)$ and radius $\mathrm{r}=5$.

Now let us find the distance d between $(-2.5,3.5)$ and $(0,0)$.

If $\mathrm{d}>\mathrm{r} \Rightarrow$ Point lies outside the circle.

If $\mathrm{d}<\mathrm{r} \Rightarrow$ Point lies inside the circle.

If $\mathrm{d}=\mathrm{r} \Rightarrow$ Point lies on the circle.

$d=\sqrt{(-2.5-0)^{2}+(3.5-0)^{2}}$

$\Rightarrow d=\sqrt{6.25+12.25}$

$\Rightarrow \mathrm{d}=\sqrt{18.5}$

$\Rightarrow \mathrm{d}=4.3$

It can be clearly seen that $4.3<5$ i.e., $d<r$.

Hence, the point lies inside the circle.

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

NCERT Solution Class 11 Maths of Chapter 11 All Exercises

Access NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections

Opting for the NCERT solutions for Ex 11.1 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.1 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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Important Topics Covered in Exercise 11.1 of Class 11 Maths NCERT Solutions

Exercise 11.1 of Class 11 Maths NCERT Solutions is mainly based on the below topics:

1. Sections of a cone

1. Circle, ellipse, parabola, and hyperbola

2. Degenerate conic sections

2. Circle

Exercise 11.1 of Class 11 Maths NCERT Solutions describes the sections of the cone with diagrams and important terms such as upper nappe, lower nappe, generator, and axis are also explained in this chapter. It explains that if the plane intersects the nappe at any point except the vertex of the cone, then the following conditions may arise :

Assume α is the angle between the axis of the cone and the rotating line and β is the angle between the intersecting plane and the vertical axis of the cone.

• If β = 90°, then the section formed is a circle.

• If α < β < 90°, then an ellipse is formed.

• If β = α, then a parabola is formed.

• If 0 ≤ β < α, then the plane cuts through both the nappes and the curves of intersection is a hyperbola.

This exercise also explains the degenerate conic sections regarding the different cases when the plane cuts at the vertex of the cone, which result in a point and straight line of a degenerate hyperbola. The students will also study an important equation of the circle. If C (h, k) is the centre, r is the radius of a circle, and P(x, y) is any point on the circle, then the equation of the circle is given by

(x – h) 2 + (y – k) 2 = r 2