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 r_{1} = \(\sqrt{4+9+12}\) = 5

r_{2} = \(\sqrt{9+81-26}\) = 8

AB = \(\sqrt{(2+3)^{2}+(3+9)^{2}}\)

= \(\sqrt{25+144}\) = 13 = r_{1} + r_{2}

∴ 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)

r_{1} = \(\sqrt{9+9-14}\) = 2,

r_{2} = \(\sqrt{1+4+4}\) = 3

AB = \(\sqrt{(-3-1)^{2}+(-3-2)^{2}}\)

= \(\sqrt{16+25}=\sqrt{41}\) > r_{1} + r_{2}

∴ 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)

r_{1} = \(\sqrt{4+1+4}\) = 3, r_{2} = \(\sqrt{1+9-6}\) = 2

AB = \(\sqrt{(2+1)^{2}+(-1-3)^{2}}\)

= \(\sqrt{9+16}\)

= 5 = r_{1} + r_{2}

∴ 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)

r_{1} = \(\sqrt{1+4+4}\) = 3, r_{2} = \(\sqrt{4+9+3}\) = 4

AB = \(\sqrt{(1+2)^{2}+(-2-3)^{2}}=\sqrt{9+16}\)

= \(\sqrt{9+25}=\sqrt{34}\) < r_{1} + r_{2}

r_{1} – r_{2} < AB < r_{2} + r_{1}

∴ 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:

c_{1} ( -3, -3) c_{2} = (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:

C_{1} (2, 1), C_{2} = (3, 2)

r_{1} = \(\sqrt{4+1-1}\) = 2

r_{2} = \(\sqrt{9+4-4}\) =3

C_{1} C_{2} = \(\sqrt{(2-3)^{2}+(1-2)^{2}}=\sqrt{2}\)

C_{1} C_{2} < r_{1} + r_{2} intersect each other 2 tangents (direct)

iii) x² + y² – 4x + 2y – 4 = 0;

x² + y² + 2x – 6y + 6 = 0.

Solution:

C_{1} (2, -1) C_{2} = (-1, 3)

C_{1} C_{2} = r_{1} + r_{2} 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:

C_{1} (-2, 3) C_{2} = (-2, 1)

r_{1} = \(\sqrt{4+9+3}\) = 4 r_{2} = \(\sqrt{4+1-4}\) = 1

C_{1} C_{2} = \(\sqrt{(-2+2)^{2}+(3-1)^{2}}\)

C_{1} C_{2} = 2 < 3 = r_{1} – r_{2}

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

C_{1} C_{2} internally in the ratio r_{1} : r_{2} = 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 C_{1} (1, 3) and C2 (0,0)

r_{1} = \(\sqrt{1+9-9}\), r_{2} = 2

External centre of similitude S¹ divides C_{1} C_{2} externally in the ratio

r_{1} : r_{2} = 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

S_{1} ≡ x² + y² – 6x – 2y + 1 =0

S_{2} ≡ x² + y² + 2x – 8y + 13 = 0

∴ The circles touch each other externally.

The point of contact P divides AB – internally in the ratio r_{1} : r_{2} = 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 S_{1} – S_{2} = 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

S_{1} ≡ x² + y² – 6x – 9y + 13 = 0

S_{2} ≡ 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 r_{1} : r_{2} = \(\frac{\sqrt{65}}{2}\) : √65

= 1 : 2 Co-ordinates of P are

P = (5, 1)

∴ Equation of the common tangent is

S_{1} – S_{2} = 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:

C_{1} = (-11, 2) C_{2} = (11, -2)

r_{1} = \(\sqrt{121+4+100}\) = 15

r_{2} = \(\sqrt{121+4-100}\) = 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:

C_{1} = (2, 5), C_{2} = (-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:

SS_{11} = S²_{1}

(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.S_{11} = S²_{1}

(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:

C_{1} = (2, -3)

r_{1} = \(\sqrt{4+9+12}\) = 5

Let C_{2} = (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:

r_{1} = \(\sqrt{4+1+4}\) = 3, r_{2} = \(\sqrt{4+1+4}\) = 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.S_{11} = S²_{1}

(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.S_{11} = S²_{1}

(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.S_{11} = S²_{1}

(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:

[x²_{1} + y²_{1} + 2gx_{1} + 2fy_{1} + c sin² α + (g² + f²) cos² α] (S)

= (xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c sin² α + (g² + f²) cos² α)²

[(-c + c sin² α)+ (g² + f²) cos² α]S

= (x (x_{1} + g) + y (y_{1} + f) + gx_{1} + fy_{1} + c sin² α + (g² + f²) cos² α)²

[cos² α (g² + f² – c)] S

= [x (x_{1} + g) + y (y_{1} + f) + gx_{1} + fy_{1} + c sin² α + (g² + f²)cos² α

Let g² + f² – c = r²

= [x (x_{1} + g) + y (y_{1} + f) + gx_{1} + fy_{1} + gx_{1} + fy_{1} + c + (cos² α).r²)²

Coefficient of x² is r² cos² α – (x_{1} + g)²

Coefficient of y² is r² cos2 α – (y_{1} + f)²

Coefficient of xy is

h. cos² α r² – 2 (x_{1} + g) (y_{1} + f)

cos θ = cos 2 α

θ = 2α

Hence proved.