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
S1 = xx1 + yy1 + g(x + x1) + f(y + y1) + 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 S1 =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 S1 = 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 – x1) (y1 + f) – (y – y1) (x1 + 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 – x1) (y1 + f) – (y – y1) (x1 + 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 (x1, 1)
P is a point on the circle
1 + 1 – 8x1 – 2 + 12 = 0
1 – 8x1 + 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, y1)
1 + y²1 = ,10 ⇒ y²1 =9
y1 = ±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 ∴ 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)