Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Circle Solutions Exercise 1(c) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Circle Solutions Exercise 1(c)

I.

Question 1.

Find the equation of the tangent at P of the circle S = 0 where P and S are given by

i) P = (7, -5), S ≡ x² + y² – 6x + 4y – 12

Solution:

Equation of the circle

S ≡ x² + y² – 6x + 4y – 12 = 0

Equation of the tangent at P(7, -5) is

S_{1} = xx_{1} + yy_{1 }+ g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x. 7 + y(-5) – 3(x + 7) + 2(y – 5) – 12 = 0

⇒ 7x – 5y – 3x – 21 + 2y – 10 – 12 = 0

4x – 3y – 43 = 0

ii) P = (-1, 1), S ≡ x² + y² – 6x + 4y – 12

Solution:

Equation of the tangent at P is

x(-1) + y . 1 – 3(x – 1) + 2(y + 1) – 12 = 0

⇒ x + y – 3x + 3 + 2y + 2 – 12 = 0

⇒ -4x + 3y – 7 = 0

⇒ 4x – 3y + 7 = 0

iii) P = (-6, -9), S ≡ x² + y² + 4x + 6y – 39

Solution:

Equation of the tangent at P is S_{1} =0

(i.e.,) x(-6) + y(-9) + 2(x – 6) + 3(y – 9) – 39 = 0

⇒ -6x – 9y + 2x – 12 + 3y – 27 – 39 = 0

⇒ – 4x- 6y- 78 = 0

⇒ 4x + 6y + 78 = 0

⇒ 2x + 3y + 39 = 0

iv) P = (3, 4), S ≡ x² + y² – 4x – 6y + 11

Solution:

Equation of the tangent at P is S_{1} = 0

⇒ x(3) + y(4) – 2 (x + 3) – 3(y + 4) + 11 =0

3x + 4y – 2x – 6 – 3y – 12 + 11 =0

x + y – 7 = 0

Question 2.

Find the equation of the normal at P of the circle S = 0 where P and S are given by

i) P = (3, -4), S ≡ x² + y² + x + y – 24

Solution:

Equation of the normal is

(x – x_{1}) (y_{1} + f) – (y – y_{1}) (x_{1} + g) = 0

(x – 3) (-4 + \(\frac{1}{2}\)) – (y + 4) (3 + \(\frac{1}{2}\)) = 0

–\(\frac{7}{2}\)(x – 3) – \(\frac{7}{2}\) (y + 4) = 0

⇒ (x – 3) + (y + 4) = 0

x – 3 + y + 4 = 0

x + y + 1 = 0

ii) P = (3, 5), S ≡ x² + y² – 10x – 2y + 6

Solution:

Equation of the normal is

(x – x_{1}) (y_{1} + f) – (y – y_{1}) (x_{1} + g) = 0

(x – 3) (5 – 1) – (y – 5) (3 – 5) = 0

4x – 12 + 2y – 10 = 0

4x + 2y – 22 = 0

or

2x + y – 11 = 0

iii) P = (1, 3), S ≡ 3(x² + y²) – 19x – 29 y + 76

Solution:

Equation of the circle is

x² + y² – \(\frac{19}{3}\)x – \(\frac{29}{3}\)y + \(\frac{76}{3}\) = 0

Equation of the normal is

(x – 1)(3 – \(\frac{29}{6}\)) – (y – 3) (1 – \(\frac{19}{6}\) ) = 0

– \(\frac{11}{6}\)(x – 1) + \(\frac{13}{6}\)(y – 3) = 0

11(x – 1) – 13(y – 3) = 0

11x – 11 – 13y + 39 = 0

11x – 13y + 28 = 0

iv) P = (1, 2), S ≡ x² + y² – 22x – 4y + 25

Solution:

Equation of the normal at P is

(x – 1) (2 – 2) – (y – 2) (1 – 11) = 0

10(y – 2) = 0 ⇒ y – 2 = 0

or y = 2

II.

Question 1.

Find the length of the chord intercepted by the circle x² + y² – x + 3y – 22 = 0 on the line y = x – 3

Solution:

Equation of the circle is

S ≡ x² + y² – x + 3y – 22 = 0

Centre C(\(\frac{1}{2}\), –\(\frac{3}{2}\))

Equation of the line is y = x – 3 ⇒ x – y – 3 = 0

P distance from the centre

\(=\frac{\left|\frac{1}{2}+\frac{3}{2}-3\right|}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}\)

Length of the chord = \(\sqrt{r^{2}-P^{2}}\)

Question 2.

Find the length of the chord intercepted by the circle x² + y² – 8x – 2y – 8 = 0 on the line x + y + 1 = 0

Solution:

Equation of the circle is x² + y² – 8x – 2y – 8 = 0

Centre is C(4, 1), r = \(\sqrt{16+1+8}\) = 5

Equation of the Sine is x + y + 1 =0

P = distance from the centre = \(\frac{|4+1+1}{\sqrt{1+1}}\)

= \(\frac{6}{\sqrt{2}}=3\sqrt{2}\)

Length of the chord = 2\(\sqrt{r^{2}-P^{2}}\)

= 2\(\sqrt{25-18}\)

= 2\(\sqrt{7}\) units.

Question 3.

Find the length of the chord formed by x² + y² = a² on the line x cos α + y sin α = p.

Solution:

Equation of the circle is x² + y² = a²

Centre C(0, 0), r = a

Equation of the line is

x cos α + y sin α – p = 0

P = distance from the centre

= \(\frac{|0+0-p|}{\sqrt{\cos ^2 \alpha+\sin ^2 \alpha}}\) = p

Length of the chord = 2\(\sqrt{r^{2}-p^{2}}\)

Question 4.

Find the equation of circle with centre (2, 3) and touching the line 3x – 4y + 1 =0.

Solution:

Equation of circle (x – h)² + (y – k)² = r²

(x – 2)² + (y – 3)² = 1

x² + y² – 4x – 6y + 12 = 0

Question 5.

Find the equation of the circle with centre (-3, 4) and touching y – axis.

Solution:

Centre of the circle is C(-3, 4)

The circle touches y – axis

r = x co – ordinates of c = |-3| = 3

Equation of the circle is (x + 3)² + (y – 4)² = 9

x² + 6x + 9 + y² – 8y + 16 – 9 = 0

x² + y² + 6x – 8y + 16 = 0

Question 6.

Find the equation of tangents of the circle x² + y² – 8x – 2y + 12 = 0 at the points whose ordinates are 1.

Solution:

Equation of the circle is

x² + y² – 8x – 2y + 12

Suppose co – ordinates of P are (x_{1}, 1)

P is a point on the circle

x²_{1} + 1 – 8x_{1} – 2 + 12 = 0

x²_{1} – 8x_{1} + 11 = 0

Co – ordinates of P are (4 + √5, 1) and Q(4 – √5, 1)

Equation of the tangent at P(4 + √5, 1) is

x(4 + √5) + y • 1 – 4(x + 4 + √5) – (y + 1) + 12 = 0

⇒ 4x + √5x + y – 4x – 16 – 4√5 – y – 1 + 12 = 0

⇒ √5x – 5 – 4√5 = 0 ⇒ √5 (x – √5 – 4) = 0

⇒ x – √5 – 4 = 0

x = 4 + √5

Equation of the tangent at Q(4 – √5, 1) is

⇒ x(4 – √5) + y. 1 – 4(x + 4 – √5) – (y + 1) + 12 = 0

⇒ 4x – √5x + y – 4x – 16 + 4√5 – y- 1 + 12 = 0

⇒ -√5x + 4√5 – 5 = 0

⇒ -√5 (x – 4 + √5) = 0

⇒ x – 4 + √5 = 0

x = 4 – √5

Question 7.

Find the equation of tangents of the circle x² + y² – 10 = 0 at the points whose abscissae are 1.

Solution:

Equation of the circle is x² + y² = 10

Suppose co-ordinates of P are (1, y_{1})

1 + y²_{1} = ,10 ⇒ y²_{1} =9

y_{1} = ±3

Co-ordinates of P are (1, 3) and (1, -3)

Equation of the tangent at P(1, 3) is

x. 1 + y. 3 = 10

x + 3y – 10 = 0

Equation of the tangent of P(1, -3) is

x. 1 + y(-3) = 10 ⇒ x – 3y – 10 = 0

III.

Question 1.

If x² + y² = c² and \(\frac{x}{a}+\frac{y}{b}\) = 1 intersect at A and B, then find \(\overline{\mathrm{AB}}\). Hence deduce the condition, that the line touches the circle.

Solution:

x² + y² = c²

C = (0, 0) r = c

Length of chord = 2\(\sqrt{r^{2}-d^{2}}\)

Line touches circle when r² = d² or r = d

Length of the chord = 0

c = \(\sqrt{\frac{a^2 b^2}{a^2+b^2}}\)

Question 2.

The line y = mx + c and the circle x² + y² = a² intersect at A and B. If AB = 2λ, then show that c² =(1 + m²) (a² – λ²)

Solution:

C = (0, 0) r = a

Length of chord 2\(\sqrt{r^{2}-d^{2}}\)

2\(\sqrt{r^{2}-d^{2}}\) = 2λ

r² – d² = λ²

y = mx + c ⇒ mx – y + c = 0

d = \(\frac{|0-0+c|}{\sqrt{m^{2}+1}}\)

d² = \(\frac{c^{2}}{m^{2}+1}\)

r² – d² = λ² and r = a

a² – d² = λ²

a² – λ² = d² = \(\frac{c^{2}}{m^{2}+1}\)

∴ c² = (a² – λ²) (1 + m²) is the required condition.

Question 3.

Find the equation of the circle with centre (-2, 3) cutting a chord length 2 units on 3x + 4y + 4 = 0

Solution:

Equation of the line is 3x + 4y + 4 = 0

P = Length of the perpendicular

Length of the chord = 2λ = 2 ⇒ λ = 1

If r is the radius of the circle then

r² = 2² + 1² – 4 + 1 = 5

Centre of the circle is (-2, 3)

Equation of the circle is (x + 2)² + (y – 3)² = 5

x² + 4x + 4 + y² – 6y + 9 – 5 = 0

i.e., x² + y² + 4x – 6y + 8 = 0

Question 4.

Find the equation of tangent and normal at (3, 2) of the circle x² + y² – x – 3y – 4 = 0.

Solution:

Equation of the circle is x² + y² – x – 3y – 4 = 0

Equation of the tangent at P (3, 2) is

x. 3 + y. 2 – \(\frac{1}{2}\)(x + 3) – \(\frac{3}{2}\)(y + 2) – 4 = 0

6x + 4y – x – 3 – 3y – 6 – 8 = 0

5x + y – 17 = 0

The normal is perpendicular to the tangent Equation of the normal can be taken as x – 5y + c = 0

The normal passes through P(3, 2)

3 – 10 + c = 0 ⇒ c = 7

Equation of the normal is x – 5y + 7 = 0

Question 5.

Find the equation of the tangent and normal at (1, 1) to the circle 2x² + 2y² – 2x – 5y + 3 = 0.

Solution:

Equation of the circle is

2x² + 2y² – 2x – 5y + 3 = 0

x² + y² – x – \(\frac{5}{2}\)y – \(\frac{3}{2}\) = 0

Equation of the tangent at P(1, 1) is

x. 1 + y.1 – \(\frac{1}{2}\)(x + 1) – \(\frac{5}{4}\)(y + 1) + \(\frac{3}{2}\) = 0

4x + 4y – 2 (x + 1) – 5(y +1) + 6 = 0

4x + 4y – 2x – 2-5y- 5 + 6 = 0

2x – y – 1 = 0

Equation of the normal can be taken as

x + 2y + k = 0

The normal passes through P(1, 1)

⇒ 1 + 2 + k = 0 ⇒ k = -3

Equation of the normal is x + 2y – 3 = 0

Question 6.

Prove that the tangent at (3, -2) of the cjrcle x² + y² = 13 touches the circle x² + y² + 2x – 10y – 26 = 0 and find its point of contact.

Solution:

Equation of the circle is x² + y² = 13

Equation of the tangent at P(3, -2) is

x. 3 + y (-2) = 13

3x – 2y – 13 = 0

Equation of the second circle is

x² + y² + 2x – 10y – 26 = 0

Centre is C(-1, 5) r = \(\sqrt{1+25+26}\)

= √52 = 2√3

P = length of the perpendicular from

= radius

∴ The tangent to the first circle also touches the second circle.

Equation of the circle is

x² + y² + 2x – 10y – 26 = 0

⇒ 3x – 2y – 13 = 0

⇒ 13x² – 130 x + 325 = 0

x² – 10x + 25 = 0 ⇒ (x – 5)² = 0

x – 5 = 0 ⇒ x = 5

x = 5, y = 1 (5, 1) is point of contact

Question 7.

Show that the tangent at (-1, 2) of the circle x² + y² – 4x – 8y + 7 = 0 touches the circle x² + y² + 4x + 6y = 0 and also find its point of contact.

Solution:

Tangent at (-1, 2) to

x² + y² – 4x – 8y + 7 = 0 is

⇒ x(-1) + y(2) – 2(x – 1) – 4(y + 2) + 7 = 0

⇒ -3x – 2y + 1 = 0 (or) 3x + 2y – 1 = 0

If 3x + 2y – 1 = 0 is tangent to

x² + y² + 4x + 6y = 0, then radius of circle should be equal to perpendicular from centre to line 3x + 2y – 1 = 0

C: (-2, -3)

d = r

Hence 3x + 2y – 1 = 0 is also tangent to x² + y² + 4x + 6y = 0

point of contact (foot of perpendicular)

Let (h, k) be foot of perpendicular

h = 1, k = -1

(1, -1) is point of contact.

Question 8.

Find the equations of the tangents to the circle x² + y² – 4x + 6y – 12 = 0 which are parallel to x + y – 8 = 0. .

Solution:

Equation of the circle is

x² + y² – 4x + 6y – 12 = 0

Centre is C(2, -3); r = radius

= \(\sqrt{4+9+12}\) = 5

Equation of the given line is x + y – 8 = 0

The tangent is parallel to this line

Equation of the tangent can be taken as x + y + k = 0

Length of the perpendicular from

c = \(\frac{|2-3+k|}{\sqrt{1+1}}\)

⇒ |k – 1| = 5√2

⇒ k – 1 = ± 5√2

⇒ k = 1 ± 5√2

Equation of the required tangent is

x + y + 1 ± 5√2 = 0

Question 9.

Find the equations of the tangents to the circle x² + y² + 2x – 2y – 3 = 0 which are perpendicular to 3x – y + 4 = 0.

Solution:

x² + y² + 2x – 2y – 3 = 0, C : (-1, 1)

r = \(\sqrt{1+1+3}\) = √5

y = –\(\frac{1}{3}\)x + k is line ⊥ to 3x – y + 4 = 0

x + 3y – 3k = 0

√5 = \(\frac{|-1+3+3k|}{\sqrt{1+9}}\)

Squaring on both sides

(3k + 2)² = 50

9k² + 12k + 4 – 50 = 0

9k² + 12k – 46 = 0

k = \(\frac{-12 \pm \sqrt{144+1656}}{18}=\frac{-12 \pm 30 \sqrt{2}}{18}\)

= \(\frac{6(-2 \pm 5 \sqrt{2})}{18}\) ⇒ 3k = -2 ± 5√2

Equation of the required tangent is

x + 3y – 2 ± 5√2 = 0

Question 10.

Find the equation of the tangents to the circle x² + y² – 4x – 6y + 3 = 0 which makes an angle 45° with X – axis.

Solution:

Equation of the circle is x² + y² – 4x – 6y + 3 = 0

Centre C(2, 3), r = \(\sqrt{4+9-3}\) = √10

Slope of the tangent m = tan 45° = 1

Equation of the tangent can be taken as

y = x + c

x – y + c = 0

Length of the perpendicular from centre =

Equation of the tangents

x – y + 1 ± 2 √5 = 0

Question 11.

Find the equation of the circle passing through (-1, 0) and touching x + y – 7 = 0 at (3, 4).

Solution:

Suppose equation of the circle is

x² + y² + 2gx + 2fy + c = 0

It passes through A (-1, 0)

1 + 0 – 2g + 0 + c = 0

2g – c = 1 ……….. (1)

Equation of the tangent at P(3, 4) is

3x + 4y + g(x + 3) + f(y + 4) + c = 0

(3 + g)x + (4 + f)y + (3g + 4f + c) = 0 ………….. (2)

Given equation of the tangent is

x + y – 7 = 0 ……….. (3)

Comparing (2) and (3)

From (1)- 2 – c = 1 ⇒ c = -3

Equation of the circle is

x² + y² – 2x – 4y – 3 = 0

Question 12.

Find the equations of the circles passing through (-1, 1), touching the lines 4x + 3y + 5 = 0 and 3x – 4y – 10 = 0.

Solution:

Suppose equation of the circle is

x² + y² + 2gx + 2fy + c = 0

Let centre be (-g, -f) 1 from centre to lines be equal (radius)

Case (i):

If g = -1; f = -2

Circle is passing through (1, -1)

∴ x² + y² + 2gx + 2fy + c = 0

1 + 1 + 2g – 2f + c = 0

2 – 2 + 4 + c = 0 (or) c = – 4

Required equation of circle be

x² + y² – 2x + 4y – 4 = 0

Case (ii):

∴ Required equation of circle be

25(x² + y²) – 26x +68y + 44 = 0

Question 13.

Show that x + y + 1 =0 touches the circle x² + y² – 3x + 7y + 14 = 0 and find its point of contact.

Solution:

x² + y² – 3x + 7y + 14 = 0

Perpendicular distance from centre on x + y + 1 = 0 if equals to radius, then x + y + 1 = 0 is tangent

∴ x + y + 1 is tangent y = -x – 1

Substituting value of ‘y’ in equation

x² + y² – 3x + 7y + 14 = 0, We get

x² + (-x – 1)² – 3x + 7 (-x – 1) + 14 = 0

⇒ 2x² – 8x + 8 = 0

⇒ x² – 4x + 4 = 0 (or) x = 2

y – x – 1, y = -3

Point of contact is (2, -3)