Class 11 Maths Ncert Solutions Chapter 11 Ex 11.1 Conic Sections PDF

# Class 11 Maths Ncert Solutions Ex 11.1

## Class 11 Maths Ncert Solutions Chapter 11 Ex 11.1

Q.1: A circle is given with centre (0, 3) and radius 3. Obtain the equation of the given circle.

Sol:

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given:

(h, k) = centre of a circle = (0, 3) and radius r = 3

The equation of a circle is:

(a – 0)2 + (b – 3)2 = 32

a2 + b2 – 6b + 9 = 9

a2 + b2 – 6b + 9 – 9 = 0

a2 + b2 – 6b = 0

Q.2: A circle is given with centre (-1, 2) and radius 4. Find the equation of the circle.

Sol:

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given:

Centre of a circle = (h, k) = (-1, 2) and radius (r) = 4

The equation of a circle is:

(a – (- 1))2 + (b – 2)2 = 42

(a + 1)2 + (b – 2)2 = 42

a2 + 2a + 1 + b2 – 4b + 4 = 16

a2 + 2a + b2 – 4b + 5 = 16

a2 + 2a + b2 – 4b + 5 – 16 = 0

a2 + 2a + b2 – 4b – 11 = 0

Q.3: A circle is given with centre (13$\frac{1}{3}$, 12$\frac{1}{2}$) and radius 114$\frac{1}{14}$. Find the equation of the circle.

Sol:

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given:

Centre of a circle = (h, k) = (13$\frac{1}{3}$, 12$\frac{1}{2}$) and the radius (r) = 4

The equation of a circle is:

(a – 13$\frac{1}{3}$)2 + (b – 12$\frac{1}{2}$)2 = 114$\frac{1}{14}$2

(a – 13$\frac{1}{3}$)2 + (b – 12$\frac{1}{2}$)2 = 114$\frac{1}{14}$2

a22.13.a+19+b22.12.b+14=(114)2a22a3+19+b2b+14=1196a2+b22a3b+19+14=1196a2+b22a3b+1336=1196a2+b22a3b+13361196=0a2+b22a3b+63791764=0a2+b22a3b+6281764=0a2+b22a3b+0.35=03a2+3b22a3b+1.068=0$a^{2} – 2.\frac{1}{3}.a + \frac{1}{9} + b^{2} – 2.\frac{1}{2}.b + \frac{1}{4} = (\frac{1}{14})^{2} \\ a^{2} – \frac{2a}{3} + \frac{1}{9} + b^{2} – b + \frac{1}{4} = \frac{1}{196} \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{1}{9} + \frac{1}{4} = \frac{1}{196} \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{13}{36} = \frac{1}{196} \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{13}{36} – \frac{1}{196} = 0 \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{637 – 9}{1764} = 0 \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{628}{1764} = 0 \\ a^{2} + b^{2} – \frac{2a}{3} – b + 0.35 = 0 \\ 3a^{2} + 3b^{2} – 2a – 3b + 1.068 = 0 \\$

Q.4: A circle is given with centre (2, 2) and radius 5$\sqrt{5}$. Obtain the equation of the given circle.

Sol:

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given:

Centre of a circle = (h, k) = (2, 2) and radius (r) = 5$\sqrt{5}$

The equation of a circle is:

(a – 2)2 + (b – 2)2 = 52$\sqrt{5}^{2}$

(a – 2)2 + (b – 2)2 = 52$\sqrt{5}^{2}$

a2 – 4a + 4 + b2 – 4b + 4 = 5

a2 – 4a + b2 – 4b + 8 = 5

a2 – 4a + b2 – 4b + 8 – 5 = 0

a2 – 4a + b2 – 4b + 3 = 0

a2 + b2 – 4a – 4b + 3 = 0

Q.5: A circle is given with centre (- x, – y) and radius x2y2$\sqrt{x^{2} – y^{2}}$. Find the equation of the circle.

Sol:

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given:

Centre of a circle = (h, k) = (- x, – y) and radius ‘r’ = x2y2$\sqrt{x^{2} – y^{2}}$

The equation of a circle is:

(a – (- x)2 + (b – (- y)2 = x2y22$\sqrt{x^{2} – y^{2}}^{2}$

(a + x)2 + (b + y)2 = x2y22$\sqrt{x^{2} – y^{2}}^{2}$

(a + x)2 + (b + y)2 = x2 – y2

a2 + 2ax + x2 + b2 + 2by + y2 = x2 – y2

a2 + 2ax + b2 + 2by + y2 = 0

Q.6: The equation of a given circle is given as (a + 5)2 + (b – 3)2 = 36. Find the centre and radius of the circle.

Sol:

Given:

The equation of a given circle (a + 5)2 + (b – 3)2 = 36

(a + 5)2 + (b – 3)2 = 36

{a – (- 5)}2 + (b  – 3) = 62

Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = – 5, k = 3 and r = 6.

Hence, the centre of the circle is (-5, 3), and the radius of circle is 6.

Q.7: The equation of a given circle is given as a2 + b2 – 4a – 4b + 3 = 0

Find the radius and centre of the circle.

Sol:

Given:

The equation of a given circle a2 + b2 – 4a – 4b + 3 = 0

a2 + b2 – 4a – 4b + 3 = 0

a2 – 4a + b2 – 4b + 3 = 0

{a2 – 2.a.2 + 22} + {b2 – 2.b.2 + 22}  – 22 – 22 + 3 = 0

{a2 – 2.a.2 + 22} + {b2 – 2.b.2 + 22}  – 8 + 3 = 0

{a2 – 2.a.2 + 22} + {b2 – 2.b.2 + 22}  = 5

(a – 2)2 + (b – 2)2 = 52$\sqrt{5}^{2}$

Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 2, k = 2 and r = 5$\sqrt{5}$.

Thus, the centre of the circle is (2, 2), and the radius of circle is 5$\sqrt{5}$.

Q.8: The equation of a given circle is given as a2 + b2 – 6a + 8b – 14 = 0

Find the centre and radius of the circle.

Sol:

Given:

The equation of a given circle a2 + b2 – 6a + 8b – 14 = 0

a2 – 6a + b2 + 8b – 14 = 0

{a2 – 2.a.3 + 32} + {b2 + 2.b.4 + 42}  – 32 – 42 – 14 = 0

(a – 3) 2 + (b + 4) 2 – 9 – 16 – 14 = 0

(a – 3) 2 + (b + 4) 2 – 39 = 0

(a – 3) 2 + (b + 4) 2 = 39

(a – 3) 2 + (b – (- 4)) 2 = 39

Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 3, k = 4 and r = 39$\sqrt{39}$.

Thus, the centre of the circle is (3, 4), and the radius of circle is 39$\sqrt{39}$.

Q.9: The equation of a given circle is given as 2a2 + 2b2 – 8a = 0

Find the centre and radius of the circle.

Sol:

Given:

The equation of a given circle 2a2 + 2b2 – 8a = 0

2a2 + 2b2 – 8a = 0

2a2 – 8a + 2b2 = 0

2[a2 – 4a + b2] = 0

{a2 – 2.a.2 + 22} – 22 + (b – 0)2 = 0

[a2 – 2.a.2 + 22] + [b – 0]2 – 22 = 0

(a – 2)2 + (b – 0)2 = 22 = 4

Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 2, k = 0 and r = 4$\sqrt{4}$ = 2.

Thus, the centre of the circle is (3, 4), and the radius of circle is 2.

Q.10: The circle passing through the points (6, 2) and (4, 3) and whose centre lies on the line 2a + y = 8. Find the equation of the circle.

Sol:

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

(a – h)2 + (b – k)2 = r2

As the circle passes through (6, 2) and (4, 3)

(6 – h)2 + (2 – k)2 = r2 . . . . . . . . . . . . . . . . (1)

(4 – h)2 + (3 – k)2 = r2 . . . . . . . . . . . . . . . . (2)

The centre (h, k) of the circle lies on the line 2a + y = 8,

2h + k = 8 . . . . . . . . . . . . . . . . . . . . . (3)

Now, On Comparing equations 1 and 2, we will get:

(6 – h)2 + (2 – k)2 = (4 – h)2 + (3 – k)2

36 – 12h + h2 + 4 – 4k + k2 = 16 – 8h + h2 + 9 – 6k + k2

36 – 12h + 4 – 4k = 16 – 8h + 9 – 6k

– 12h – 4k + 8h + 6k = 16 + 9 – 36 – 4

4h – 2k = 15 . . . . . . . . . . . . . . . . . . (4)

Now, on Solving Equations (3) and (4), we will get:

h = 318$\frac{31}{8}$

k = 14$\frac{1}{4}$

Substituting the values of h and k in equation (1), we will get:

(6 – 318$\frac{31}{8}$)2 + (2 – 14$\frac{1}{4}$)2 = r2

(178)2+(74)2=r2$\\(\frac{17}{8})^{2} + (\frac{7}{4})^{2} = r^{2}$

$\\\Rightarrow$ 28964+4916=r2$\frac{289}{64} + \frac{49}{16} = r^{2}$

$\\\Rightarrow$  289+19664=r2$\frac{289 + 196}{64} = r^{2}$

$\\\Rightarrow$ 48564=r2$\frac{485}{64} = r^{2}$

$\Rightarrow$  r = 7.57

The required equation is:

(a – 318$\frac{31}{8}$)2 + (b – 14$\frac{1}{4}$)2 = r2

(a – 318$\frac{31}{8}$)2 + (b – 14$\frac{1}{4}$)2 = 7.572

(a318)2+(b14)2=7.572$(a – \frac{31}{8})^{2} + (b – \frac{1}{4})^{2} = 7.57^{2} \\$

$\\\Rightarrow$   a2314a+(318)2+b2b2+116=57.30$a^{2} – \frac{31}{4}a + (\frac{31}{8})^{2} + b^{2} – \frac{b}{2} + \frac{1}{16} = 57.30\\$

$\\\Rightarrow$   a2314a+96164+b2b2+116=57.30$a^{2} – \frac{31}{4}a + \frac{961}{64} + b^{2} – \frac{b}{2} + \frac{1}{16} =57.30$

Therefore, the required equation is: 64a2– 496a + 64b2 -32b = 2702.51

Q.11: The circle passing through the points (3, 2) and (-2, 2) and whose centre lies on the line a – 3y – 13 = 0. Find the equation of the circle.

Sol:

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

As the circle passes through (3, 2) and (-2, 1)

(3 – h)2 + (2 – k)2 = r2 . . . . . . . . . . . . . . . . (1)

(-2 – h)2 + (2 – k)2 = r2 . . . . . . . . . . . . . . . (2)

The centre (h, k) of the circle lies on the line a – 3y – 13 = 0,

h – 3k = 13 . . . . . . . . . . . . . . . (3)

Now, on comparing equations 1 and 2, we will get:

(3 – h)2 + (2 – k)2 = (-2 – h)2 + (2 – k)2

9 – 6h + h2 = 4 + 4h + h2

9 – 6h = 4 + 4h

h = 0.5 . . . . . . . . . . . . . . . . . . (4)

Now, on solving equation (3), we will get:

k = – 256$\frac{25}{6}$ = – 4.166

Now, on substituting the values of h and k in equation (1), we will get:

(3 – 0.5)2 + (2 – (- 256$\frac{25}{6}$))2 = r2

(2.5) 2 + (2 + 256$\frac{25}{6}$)2 = r2

(2.5) 2 + (2 + 4.16)2 = r2

(2.5) 2 + (6.16)2 = r2

r2 = 44.1956

r = 6.65

The required equation is:

(a – h)2 + (b – k)2 = r2

(a – 0.5)2 + (b – (- 4.16))2 = r2

(a – 0.5)2 + (b + 4.16)2 = r2

a2 – a + 6.25 + b2 + 8.32b + 17.30 = 44.19

a2 – a + b2 + 8.32b = 44.19 – 6.25 – 17.30

a2 + b2 – a + 8.32b – 20.64 = 0

Q.12: The radius of the circle is 6 whose centre lies on x – axis and passes through the point (3, 2). Find the equation.

Sol:

Given:

The radius of the circle is 6 and passes through the point (3, 2)

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

As, the centre lies on x – axis, k = 0, the equation becomes:

(a – h) 2 + b 2 = 36

(3 – h) 2 + 22 = 36

(3 – h) 2 = 36 – 4

3 – h = 32$\sqrt{32}$

h = 3+42$3 + 4\sqrt{2}$ or h = 342$3 – 4\sqrt{2}$

Q.13: The circle passing through (0, 0) and making intercepts a and b on the coordinate axes. Find the equation of the circle.

Sol:

Suppose, the equation of the required circle be (x – h) 2 + (y – k) 2 = r 2

The centre of the circle passes through (0, 0):

(0 – h) 2 + (0 – k) 2 = r 2

h 2 + k 2 = r 2

Therefore, the equation of the circle is:

(x – h) 2 + (y – k) 2 = h 2 + k 2 .

Given, the circle makes intercepts a and b on the coordinate axes.

The circle passes through points (a, 0) and (0, b).

Therefore, (a – h) 2 + (0 – k) 2 = h 2 + k 2  . . . . . . . . . . . . . . . (1)

(0 – h) 2 + (b – k) 2 = h 2 + k 2 . . . . . . . . . . . . . . . . . . (2)

Now, from equation (1), we obtain:

a 2 – 2ah + h 2 + k 2 = h 2 + k 2

a 2 – 2ah = 0

a (a – 2h) = 0

a = 0 or (a – 2h) = 0

We know, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =a/2.

Now, from equation (2), we will get:

h 2 + b 2 – 2bk + k 2 = h 2 + k 2

b 2 – 2bk = 0

b (b – 2k) = 0

b = 0 or(b – 2k) = 0

However, b ≠ 0;

Therefore, (b – 2k) = 0 ⇒ k =b/2.

Thus, the equation of the required circle is:

(xa2)2+(yb2)2=(a2)2+(b2)2(2xa2)2+(2yb2)2=a2+b224x24ax+a2+4y24bx+b2=a2+b24x2+4y24ax4by=0x2+y2axby=0$(x – \frac{a}{2})^{2} + (y – \frac{b}{2})^{2} = (\frac{a}{2})^{2} + (\frac{b}{2})^{2} \\ (\frac{2x – a}{2})^{2} + (\frac{2y – b}{2})^{2} = \frac{a^{2} + b^{2}}{2} \\ 4x^{2} – 4ax + a^{2} + 4y^{2} – 4bx + b^{2} = a^{2} + b^{2} \\ 4x^{2} + 4y^{2} – 4ax – 4by = 0 \\ x^{2} + y^{2} – ax – by = 0$

Q.14: The circle with centre (3, 3) and passes through the point (6, 5), find the equation of a circle?

Sol:

Given:

The centre of the circle (h, k) = (3, 3).

As the circle passes through point (6, 5), the radius (r) of the circle is the distance between the points (2, 2) and (6, 5).

r=(36)2+(35)2+(3)2+(2)2=9+4=13$r = \sqrt{(3 – 6)^{2} + (3 – 5)^{2}} + \sqrt{(- 3)^{2} + (- 2)^{2}} = \sqrt{9 + 4} = \sqrt{13}\\$

Therefore, the equation of the circle is:

(ah)2+(bk)2=r2(a3)2+(b3)2=132a26a+9+b26b+9=13a2+b26a6b+1813=0a2+b26a6b+5=0$\\(a – h)^{2} + (b – k)^{2} = r^{2} \\ (a – 3)^{2} + (b – 3)^{2} = \sqrt{13}^{2} \\ a^{2} – 6a + 9 + b^{2} – 6b + 9 = 13 \\ a^{2} + b^{2} – 6a – 6b + 18 – 13 = 0 \\ a^{2} + b^{2} – 6a – 6b + 5 = 0$

Q.15: Check whether the point (–2.0, 3.0) lies inside, outside or on the circle x 2 + y 2 = 25?

Sol:

Given:

The equation of the given circle is x 2 + y 2 = 25

x 2 + y 2 = 25

(x – 0)2 + (y – 0)2 = 52 , which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 0, k = 0, and r = 5.

Centre = (0, 0) and radius = 5

Distance between point (– 2.0, 3.0) and centre (0, 0)

(20)2+(30)2$\sqrt{(- 2 – 0)^{2} + (3 – 0)^{2}}\\$

=4+9=13=3.60(approx.)<5$\sqrt{4 + 9} = \sqrt{13} = 3.60 (approx.) < 5$

As, the distance between point (– 2.0, 3.0) and centre (0, 0) of the circle is less than the radius of the circle, point (– 2.0, 3.0) lies inside the circle.