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

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

I.

Question 1.
Discuss the relative position of the following pair of circles.
i) x² + y² – 4x – 6y – 12 = 0
x² + y² + 6x + 18y + 26 = 0.
Solution:
Centres of the circles are A (2, 3), B(-3, -9)
radii are r1 = $$\sqrt{4+9+12}$$ = 5
r2 = $$\sqrt{9+81-26}$$ = 8
AB = $$\sqrt{(2+3)^{2}+(3+9)^{2}}$$
= $$\sqrt{25+144}$$ = 13 = r1 + r2
∴ The circles touch externally.

ii) x² + y² + 6x + 6y + 14 = 0,
x² + y² – 2x – 4y – 4 = 0.
Solution:
Centres are A (-3, -3), B (1, 2)
r1 = $$\sqrt{9+9-14}$$ = 2,
r2 = $$\sqrt{1+4+4}$$ = 3
AB = $$\sqrt{(-3-1)^{2}+(-3-2)^{2}}$$
= $$\sqrt{16+25}=\sqrt{41}$$ > r1 + r2
∴ Each circle lies exterior to the other circle.

iii) (x – 2)² + (y + 1)² = 9, (x + 1)² + (y – 3)²= 4
Solution:
Centres are A( 2, -1), B(-1, 3)
r1 = $$\sqrt{4+1+4}$$ = 3, r2 = $$\sqrt{1+9-6}$$ = 2
AB = $$\sqrt{(2+1)^{2}+(-1-3)^{2}}$$
= $$\sqrt{9+16}$$
= 5 = r1 + r2
∴ The circles touch each other externally.

iv) x² + y² – 2x + 4y – 4 = 0, x² + y² + 4x – 6y – 3 = 0
Solution:
Centres are A (1, -2), B (-2, 3)
r1 = $$\sqrt{1+4+4}$$ = 3, r2 = $$\sqrt{4+9+3}$$ = 4
AB = $$\sqrt{(1+2)^{2}+(-2-3)^{2}}=\sqrt{9+16}$$
= $$\sqrt{9+25}=\sqrt{34}$$ < r1 + r2
r1 – r2 < AB < r2 + r1
∴ The circles intersect each other.

Question 2.
Find the number of possible common tangents that exist for the following pairs of circles.
i) x² + y² + 6x + 6y + 14 = 0,
x² + y² – 2x – 4y – 4 = 0
Solution:
c1 ( -3, -3) c2 = (1, 2)

The circles are externally No. of common tangents = 4

ii) x² + y² – 4x – 2y + 1 = 0;
x² + y² – 6x – 4y + 4 = 0.
Solution:
C1 (2, 1), C2 = (3, 2)
r1 = $$\sqrt{4+1-1}$$ = 2
r2 = $$\sqrt{9+4-4}$$ =3

C1 C2 = $$\sqrt{(2-3)^{2}+(1-2)^{2}}=\sqrt{2}$$
C1 C2 < r1 + r2 intersect each other 2 tangents (direct)

iii) x² + y² – 4x + 2y – 4 = 0;
x² + y² + 2x – 6y + 6 = 0.
Solution:
C1 (2, -1) C2 = (-1, 3)

C1 C2 = r1 + r2 touch each other externally; No. of common tangents = 3.

iv) x² + y² = 4; x² + y² – 6x – 8y + 16 = 0
Solution:

Two circles touch each other.
2 – direct tangents
1 – transverse tangent (at common point)

v) x² + y² + 4x – 6y – 3 = 0
x² + y² + 4x – 2y + 4 = 0.
Solution:
C1 (-2, 3) C2 = (-2, 1)
r1 = $$\sqrt{4+9+3}$$ = 4 r2 = $$\sqrt{4+1-4}$$ = 1
C1 C2 = $$\sqrt{(-2+2)^{2}+(3-1)^{2}}$$
C1 C2 = 2 < 3 = r1 – r2
One circle is inside the other.
∴ No common tangent.

Question 3.
Find the internal centre of similitude for the circles x² + y² + 6x – 2y + 1 =0 and x² + y² – 2x – 6y + 9 = 0.
Solution:

The internal centre of similitude S divides
C1 C2 internally in the ratio r1 : r2 = 3 : 1
Co-ordinates of S¹ are

Question 4.
Find the external centre of similitude for the circles x² + y² – 2x – 6y + 9 = 0 and x² + y² = 4.
Solution:
Centre of the circle C1 (1, 3) and C2 (0,0)
r1 = $$\sqrt{1+9-9}$$, r2 = 2
External centre of similitude S¹ divides C1 C2 externally in the ratio
r1 : r2 = 1 : 2

II.

Question 1.
i) Show that the circles x² + y² – 6x – 2y + 1 = 0 ; x² + y² + 2x – 8y + 13 = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.
Solution:
Equations of the circles are
S1 ≡ x² + y² – 6x – 2y + 1 =0
S2 ≡ x² + y² + 2x – 8y + 13 = 0

∴ The circles touch each other externally.
The point of contact P divides AB – internally in the ratio r1 : r2 = 3 : 2
Co-ordinates of P are
($$\frac{3(-1)+2.3}{5}$$, $$\frac{3.4+2.1}{5}$$ i.e., P($$\frac{3}{5}$$, $$\frac{14}{5}$$)
Equation of the common tangent is S1 – S2 = 0
-8x + 6y – 12 = 0 (or) 4x – 3y + 6 = 0

ii) Show that x² + y² – 6x – 9y + 13 = 0, x² + y² – 2x – 16y = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.
Solution:
Equations of the circles are
S1 ≡ x² + y² – 6x – 9y + 13 = 0
S2 ≡ x² + y² – 2x – 16y = 0
centres are A (3, $$\frac{9}{2}$$), B(1, 8)

∴ The circles touch each other internally. The point of contact ‘P’ divides AB externally in the ratio r1 : r2 = $$\frac{\sqrt{65}}{2}$$ : √65
= 1 : 2 Co-ordinates of P are

P = (5, 1)
∴ Equation of the common tangent is
S1 – S2 = 0
– 4x + 7y +13 = 0
4x – 7y – 13 = 0

Question 2.
Find the equation of the circle which touches the circle x² + y² – 2x – 4y – 20 = 0 externally at (5, 5) with radius 5.
Solution:
x² + y² – 2x – 4y – 20 = 0 ,
C = (1, 2)

Equation of circle be
(x – 9)² + (y – 8)²
x² + y² – 18x – 16y + 120 = 0
If (h, k) is the centre of the required circle (5, 5) is the mid-point of (1, 2) and (h, k).

Question 3.
Find the direct common tangents of the circles.
x² + y² + 22x – 4y – 100 = 0 and x² + y² – 22x + 4y + 100 = 0.
Solution:
C1 = (-11, 2) C2 = (11, -2)
r1 = $$\sqrt{121+4+100}$$ = 15
r2 = $$\sqrt{121+4-100}$$ = 5
Let y = mx + c be tangent
mx – y + c = 0
⊥ from (-11, 2) to tangent = 15
⊥ from (11 ,-2) to tangent = 5

c = -22 m – 4
∴ y = mx – 22m – 4
This line is a tangent to the second circle
∴ $$\frac{|11m+2-22m-4|}{\sqrt{m^{2}+1}}$$ = 5
Squaring and cross multiplying
25(1 + m²) = (11m + 2 – 22m – 4)²
96m² + 44m – 21 = 0
⇒ 96m² + 72m – 28m – 21 = 0

⇒ 24y = 7x – 250
⇒ 7x – 24y – 250 = 0

Question 4.
Find the transverse common tangents of the circles x² + y² – 4x – 10y + 28 = 0 and x² + y² + 4x – 6y + 4 = 0.
Solution:
C1 = (2, 5), C2 = (-2, 3)

= [$$\frac{4}{4}$$, $$\frac{9}{2}$$] = [1, $$\frac{9}{2}$$]
Equation of the pair transverse of the common tangents is

Question 5.
Find the pair of tangents from (4, 10) to the circle x² + y² = 25.
Solution:
SS11 = S²1
(x² + y² – 25) (16 + 100 – 25)
= (4x + 10 y -25)²
91x² + 91y² – 2275
= [16x² + 100y² + 625 + 80xy – 200x – 500y]
75x² – 9y² – 80xy + 500y + 200x – 2900 = 0

Question 6.
Find the pair of tangents drawn from (0,0) to x2 + y2 + 10x + 10y + 40 = 0.
Solution:
S.S11 = S²1
(x² + y² + 10x + 10y + 40) (40) = [5x + 5y + 40]²
8(x² + y² + 10x + 10y + 40) = (x + y + 8)²5
8x² + 8y² + 80x + 80y + 320
= 5x² + 5y² + 10xy + 80x + 80y + 320
3x² + 3y² – 10xy = 0

III.

Question 1.
Find the equation of the circle which touches x² + y² – 4x + 6y – 12 = 0 at (-1, 1) internally with a radius of 2.
Solution:
C1 = (2, -3)
r1 = $$\sqrt{4+9+12}$$ = 5

Let C2 = (h, k)
point of contact – (x, y) = (-1, 1)
since the two circles touch internally

Equation of a circle with centre ($$\frac{1}{5}$$, $$\frac{-3}{5}$$) and radius 2 is given by
(x – $$\frac{1}{5}$$)² + (y + $$\frac{1}{5}$$)² = 4
5x² + 5y² – 2x + 6y – 18 = 0

Question 2.
Find all common tangents of the following pairs of circles.
(i) x² + y² = 9 and x² + y² – 16x + 2y + 49 = 0
Solution:
The circles are x² + y² = 9
x² + y² – 16x + 2y + 49 = 0
Centre are A (0, 0), B(8, -1)

The circles lie outside each other.
A (0, 0), B (8, -1)
External center of similitude S divides AB externally in this ratio 3 : 4
Co-ordinates of are (-24, + 3)
Suppose m is the slope of the direct common tangents
y – 3 = m(x + 24)
= mx + 24m
mx – y 4- (24m + 3) = 0 ………… (1)
This is a tangent to the circle x² + y² = 9
3 = $$\frac{|24m+3|}{\sqrt{m^{2}+1}}$$
9(m² +1 = 9(m² + 1) = 9(8m + 1)²
= 64m² + 16m + 1
63m² + 16m = 0
m(63m + 16) = 0
m = 0 or $$\frac{-16}{63}$$.

Case (i): m = 0
Substituting in (1), equation of the tangent is
-y + 3 = 0
y – 3 = 0

Case (ii) : m = $$\frac{-16}{63}$$.
Equation of the tangent is

Internal center of similitude S’ divides AB internally in the ratio 3 : 4
Co-ordinates of S’ are ($$\frac{24}{7}$$, $$\frac{-3}{7}$$)
Equation of the transverse common tangent is

7y + 3 = 7mx – 24m
7mx – 7y – (24m + 3) = 0 ………… (2)
This is a tangent to the circle x² + y² = 9
3 = $$\frac{|24m+3|}{\sqrt{49m^{2}+49}}=\frac{3}{\sqrt{7}} \frac{|28m+1|}{\sqrt{m^{2}+1}}$$
49 (m² + 1) = (8m + 1)²
49m² + 49 = 64m² +16m + 1
15m² + 16m – 48 = 0
(3m – 4) (5m + 12) = 0
m = $$\frac{4}{3}$$ or $$\frac{-12}{5}$$

Case (i): Substituting in (2), equation of the tangent is

Case (ii): m = $$\frac{-12}{5}$$
Equation of the transverse common tangent is ie

∴ Equation of direct common tangents are
y – 3 = 0; 16x + 63y + 195 = 0
Equation of transverse common tangents are
4x – 3y – 15 = 0 and 12x + 5y – 39 = 0

ii) x² + y² + 4x + 2y – 4 = 0 and x² + y² – 4x – 2y + 4 = 0
Solution:
r1 = $$\sqrt{4+1+4}$$ = 3, r2 = $$\sqrt{4+1+4}$$ = 1
External center of similitude S divides AB externally in the ratio 3 : 1
3 : 1 A (-2,-1), S’ B (2, 1) S
Co-ordinates of S are
($$\frac{3.2-(-2)1}{3-1}$$, $$\frac{3.1-1(-1)}{3-1}$$) = (4, 2)
Suppose m is the slope of the tangent Equation of the tangent can be taken as
y – 2 = m(x – 4)
= mx – 4m
mx – y + (2 – 4m) = 0 ………… (1)
This is a tangent to the circle
x² + y² + 4x + 2y – 4 = 0

Squaring and cross-multiplication is
(1 – 2m)² = (m² + 1)
4m² – 4m + 1 = m² + 1
3m² – 4m = 0
m(3m-4) = 0
m,= 0 or = $$\frac{4}{3}$$
m = 0
Substituting in (1), equation of the tangent is y + 2 = 0 or y -2 = 0
m = $$\frac{4}{3}$$
Substituting in (1), equation of the tangent is
$$\frac{4}{3}$$x – y + (2 – $$\frac{16}{3}$$) = 0
⇒ $$\frac{4}{3}$$x – y –$$\frac{10}{3}$$ = 0
4x – 3y – 10 = 0
Internal centre of similitude S’ divides AB internally in the ratio 3 : 1
Co-ordinates of S’ are
($$\frac{6-2}{3+1}$$, $$\frac{3-1}{3+1}$$x) = (1, $$\frac{1}{2}$$)
Equation of the tangent can be taken as
y – $$\frac{1}{2}$$ = m(x – 1)
= mx – m
mx – y + ($$\frac{1}{2}$$ – m) = 0 ………. (2)
This is a tangent to the circle
x² + y² + 4x + 2y – 4 = 0

Squaring and cross-multiply is
(1 – 2m)² = 4(m² + 1)
1 + 4m² – 4m = 4m² + 4
Here then value of m is a, so that the tangent is a vertical line, equation of the tangent is
x = $$\frac{1}{2}$$
x – 1 = 0
4m + 3 = 0
m = $$\frac{-3}{4}$$
Substituting in (2), equation of the tangent is
$$\frac{-3}{4}$$x – y + ($$\frac{1}{2}+\frac{3}{4}$$) = 0
-3x – 4y + 5 = 0
3x + 4y – 5 = 0
Equation of direct common tangents are
y – 2 = 0, 4x – 3y – 10 = 0
Equation of transvere common tangents are
x – 1 = 0, 3x + 4y – 5 = 0

Question 3.
Find the pair of tangents drawn from (3, 2) to the circle x² + y² – 6x + 4y – 2 = 0.
Solution:
S.S11 = S²1
(x² + y² – 6x + 4y – 2)
(9+ 4 – 6 × 3 + 4 × 2 – 2)
= (3x + 2y – 3(x + 3) + 2(y + 2) – 2)²
(x² + y² – 6x + 4y – 2) = (4y – 7)²
x² + y² – 6x + 4y – 2 = 16y² – 56y + 49
x² – 15y² – 6x + 60y – 51 = 0

Question 4.
Find the pair of tangents drawn from (1, 3) to the circle x² + y² – 2x + 4y – 11 = 0 and also find the angle between them.
Solution:
S.S11 = S²1
(x² + y² – 2x + 4y -11) (1 + 9-2 + 12-11) = [x + 3y – 1 (x + 1) + 2 (y + 3)- 11]²
(x² + y² – 2x + 4y – 11)9 = [5y-6]²
9x² + 9y² – 18x + 36y – 99
25y²+ 36 – 60 y
9x² – 16y² – 18x + 96y – 135 = 0

Question 5.
Find the pair of tangents from the origin to the circle x² + y² + 2gx + 2fy + c = 0 and hence deduce a condition for these tangents to be perpendicular.
Solution:
S.S11 = S²1
(x² + y² + 2gx + 2fy + c) (c) = [gx + fy + c]²
= g²x² +f²y²+ 2gfxy + 2gcx + 2fyc + c²
(gx + fy)² = c(x² + y²)
g²x² + f²y² + 2fg xy = cx² + cy²
(g² – c)x + 2fgxy + (f² – c)y² = 0
co-efficient of x² + co-efficient of y² = 0
g² – c + f² – c = 0
g² + f² = 2c

Question 6.
From a point on the circle x² + y² + 2gx + 2fy + c = 0, two tangents are drawn to the circle x² + y² + 2gx + 2fy + c sin² α + (g² + f²) cos² α = 0 (0 < a < π/2). Prove that the angle between them is 2α.
Solution:
[x²1 + y²1 + 2gx1 + 2fy1 + c sin² α + (g² + f²) cos² α] (S)
= (xx1 + yy1 + g(x + x1) + f(y + y1) + c sin² α + (g² + f²) cos² α)²
[(-c + c sin² α)+ (g² + f²) cos² α]S
= (x (x1 + g) + y (y1 + f) + gx1 + fy1 + c sin² α + (g² + f²) cos² α)²
[cos² α (g² + f² – c)] S
= [x (x1 + g) + y (y1 + f) + gx1 + fy1 + c sin² α + (g² + f²)cos² α
Let g² + f² – c = r²
= [x (x1 + g) + y (y1 + f) + gx1 + fy1 + gx1 + fy1 + c + (cos² α).r²)²
Coefficient of x² is r² cos² α – (x1 + g)²
Coefficient of y² is r² cos2 α – (y1 + f)²
Coefficient of xy is
h. cos² α r² – 2 (x1 + g) (y1 + f)

cos θ = cos 2 α
θ = 2α
Hence proved.