Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.1

In each of the following questions 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. Centre (\(\frac{1}{2}\), \(\frac{1}{4}\) and radius = \(\frac{1}{12}\).

4. Centre (1, 1) and radius \(\sqrt{2}\).

5. Centre (- a, – b) and radius \(\sqrt{a^{2}-b^{2}}\).

Solutions to questions 1 to 5:

1. Centre of the circle (0, 2) and radius = 2.

∴ Equation of the circle is

(x – 0)^{2} + (y – 2)^{2} = 2^{2}.

or x^{2} + y^{2} – 4y = 0.

2. Centre of the circle is (- 2, 3) and radius is 4.

∴ Equation of the circle is

(x + 2)^{2} + (y – 3)^{2} = 4^{2}.

or x^{2} + y^{2} + 4x – 6y + 4 + 9 – 16 = 0 or x^{2} + y^{2} + 4x – 6y – 3 = 0.

3. Centre of circle is (\(\frac{1}{2}\), \(\frac{1}{4}\) and radius is \(\frac{1}{12}\).

∴ Equation of the circle is

(x – \(\frac{1}{2}\))^{2} + (y – \(\frac{1}{4}\))^{2} = \(\frac{1}{12^{2}}\) = \(\frac{1}{144}\).

or 144x^{2} + 144y^{2} – 144x – 72y + 45 = 1.

or 144x^{2} = 144y^{2} – 144x – 72y + 44 = 0.

or 36x^{2} + 36y^{2} – 36x – 18y + 11 = 0.

4. Centre of circle is (1, 1) and radius = \(\sqrt{2}\)

∴ Equation of circle is

(x – 1)^{2} + (y – 1)^{2} = (\(\sqrt{2}\))^{2} = 2

or x^{2} + y^{2} – 2x – 2y + 2 = 2

or x^{2} + y^{2} – 2x – 2y = 0.

5. Centre of the circle is (- a, – b) and radius = \(\sqrt{a^{2}-b^{2}}\)

∴ Equation of circle is

(x + a)^{2} + (y + b)^{2} = a^{2} – b^{2}.

or x^{2} + y^{2} + 2xa + 2yb + a^{2} + b^{2} = a^{2} – b^{2}.

or x^{2} + y^{2} + 2ax + 2by + 2b^{2} = 0.

In each of the following questions 6 to 9, find the centre and radius of the circle:

6. (x + 5)^{2} + (y – 3)^{2} = 36

7. x^{2} + y^{2} – 4x – 8y – 45 = 0

8. x^{2} + y^{2} – 8x + 10y – 12 = 0

9. 2x^{2} + 2y^{2} – x = 0.

Solutions to questions 6 to 9:

6. Comparing the equation of the circle

(x + 5)^{2} + (y – 3)^{2} = 36

with (x – h)^{2} + (y – k)^{2} = r^{2}, we get

– h = 5 or h = – 5, k = 3, r^{2} = 36 or r = 6.

∴ Centre of the circle is (- 5, 3) and radius = 6.

7. Equation of the circle is

x^{2} + y^{2} – 4x – 8y = 45.

or (x^{2} – 4x) + (y^{2} – 8y) = 45

or (x^{2} – 4x + 4) + (y^{2} – 8y + 16) = 45 + 4 + 16 = 65.

⇒ (x – 2)^{2} + (y – 4)^{2} = 65

∴ Centre is (2, 4) and radius = \(\sqrt{65}\).

8. The equation of the circle is

x^{2} + y^{2} – 8x + 10y – 12

or (x^{2} – 8x) + (y^{2} + 10y) = 12.

or (x^{2} – 8x + 16) + (y^{2} + 10y + 25) = 12 + 16 + 25.

or (x – 4)^{2} + (y + 5)^{2} = 53.

∴ Centre is (4, – 5) and radius is \(\sqrt{53}\).

9. Equation of the circle is

2x^{2} + 2y^{2} – x = 0

or x^{2} + y^{2} – \(\frac{x}{2}\) = 0.

∴ Centre is (\(\frac{1}{4}\), 0) and radius = \(\frac{1}{4}\).

Question 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 circle be

(x – h)^{2} + (y – k)^{2} = r^{2} ………………. (1)

The points (4, 1) and (6, 5) lies on it.

∴ (4 – h)^{2} + (1 – k)^{2} = r^{2}

or h^{2} + k^{2} – 8h – 2k + 17 = r^{2} …………….. (2)

and (6 – h)^{2} + (5 – k)^{2} = r^{2}

or h^{2} + k^{2} – 12h – 10k + 61 = r^{2} ………………….. (3)

The centre (h, k) lies on

4x + y = 16.

∴ 4h + k = 16 ………………….. (4)

Subtracting (3) from (2), we get

4h + 8k – 44

or h + 2k = 11 ……………………. (5)

Multiplying (5) by 4, we get

4h + 8k = 44.

Subtracting equation (4) from it, we get

7k = 44 – 16 = 28. ⇒ k = 4.

From (5), h + 8 = 11. ⇒ h = 3.

Putting h = 3, k = 4 in (2), we get

9 + 16 – 24 – 8 + 17 = r^{2}

or 42 – 32 = r^{2}

∴ r^{2} = 10.

∴ Equation of the circle is

(x – 3)^{2} + (y – 4)^{2} = 10

or x^{2} + y^{2} – 6x – 8y + 15 = 0.

Question 11.

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.

Solution:

Let the equation of the circle be

(x – h)^{2} + (y – k)^{2} = r^{2} ………………. (1)

The points (2, 3) and (- 1, 1) lies on it.

∴ (2 – h)^{2} + (3 – k)^{2} = r^{2}

or h^{2} + k^{2} – 4h – 6k + 13 = r^{2} …………………… (2)

and (- 1 – h)^{2} + (1 – k)^{2} = r^{2} …………………. (3)

Centre (h, k) lies on x – 3y – 11 = 0.

∴ h – 3k – 11 = 0 …………….. (4)

Subtracting (2) from (3), we get

6h + 4k – 11 = 0 ………………. (5)

Multiplying (4) by 6, we get

6h – 18k – 66 = 0 …………………. (6)

Subtracting (6) from (5), we get

22k+ 55 = 0.

∴ k = – \(\frac{55}{22}\) = – \(\frac{5}{2}\).

From (4), h = 3k + 11 = – \(\frac{15}{2}\) + 11 = \(\frac{7}{2}\).

Putting the values of h and k in (2 – h)^{2} + (3 – k)^{2} = r^{2}, we get

or x^{2} + y^{2} – 7x + 5y – 14 = 0.

Question 12.

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

Solution:

Let the equation of the circle be

(x – h)^{2} + (y – k)^{2} = r^{2} ……………………. (1)

Here, r = 5, So, r^{2} = 25.

Centre lies onx-axis, i.e.,k = 0.

∴ Eq. (1) becomes

(x – h)^{2} + y^{2} = 25.

(2, 3) lies on it.

∴ (2 – h)^{2} + 9 = 25

or (2 – h)^{2} = 16

∴ 2 – h = ± 4

∴ h = – 2, 6

When h = – 2, equation of circle is

(x + 2)^{2} + y^{2} = 25

or x^{2} + y^{2} + 4x – 21 = 0.

When h = 6, equation of the circle is

(x – 6)^{2} + y^{2} = 25

x^{2} + y^{2} – 12x + 11 = 0.

Thus, required circles are

x^{2} + y^{2} + 4x – 21 = 0 and x^{2} + y^{2} – 12x + 11 = 0.

Question 13.

Find the equation of the circle passing through (0, 0) and making intercepts a and b on the co-ordinate axes.

Solution:

a and b are the intercepts made by the circle on the co-ordinate axes at A and B. C, the mid-point of AB is the centre of the circle.

∴ Centre (\(\frac{a}{2}\), \(\frac{b}{2}\)).

Question 14.

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

Solution:

Centre of the circle C(2, 2).

P(4, 5) is a point on the circle.

∴ Radius = CP = \(\sqrt{(4-2)^{2}+(5-2)^{2}}\)

= \(\sqrt{4+9}\) = \(\sqrt{13}\).

∴ Equation of the circle is

(x – 2)^{2} + (y – 2)^{2} = 13.

or x^{2} + y^{2} – 4x – 4y = 5.

Question 15.

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

Solution:

Centre C of the circle is (0, 0) and radius = 5.

The point is P(- 2.5, 3.5).

∴ CP < Radius.

∴ P lies inside the circle.