Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Miscellaneous Exercise 6 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Miscellaneous Exercise 6

(I) Choose the correct alternative.

Question 1.

Equation of a circle which passes through (3, 6) and touches the axes is

(A) x^{2} + y^{2} + 6x + 6y + 3 = 0

(B) x^{2} + y^{2} – 6x – 6y – 9 = 0

(C) x^{2} + y^{2} – 6x – 6y + 9 = 0

(D) x^{2} + y^{2} – 6x + 6y – 3 = 0

Answer:

(C) x^{2} + y^{2} – 6x – 6y + 9 = 0

Question 2.

If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 sq. units, then find the equation of the circle.

(A) x^{2} + y^{2} – 2x + 2y = 40

(B) x^{2} + y^{2} – 2x – 2y = 47

(C) x^{2} + y^{2} – 2x + 2y = 47

(D) x^{2} + y^{2} – 2x – 2y = 40

Answer:

(C) x^{2} + y^{2} – 2x + 2y = 47

Hint:

Centre of circle = Point of intersection of diameters.

Solving 2x – 3y = 5 and 3x – 4y = 7, we get

x = 1, y = -1

Centre of the circle C(h, k) = C(1, -1)

∴ Area = 154

πr^{2} = 154

\(\frac{22}{7} \times r^{2}\) = 154

r^{2} = 154 × \(\frac{22}{7}\) = 49

∴ r = 7

equation of the circle is

(x – 1)^{2} + (y + 1)^{2} = 72

x^{2} + y^{2} – 2x + 2y = 47

Question 3.

Find the equation of the circle which passes through the points (2, 3) and (4, 5), and the center lies on the straight line y – 4x + 3 = 0.

(A) x^{2} + y^{2} – 4x – 10y + 25 = 0

(B) x^{2} + y^{2} – 4x – 10y – 25 = 0

(C) x^{2} + y^{2} – 4x + 10y – 25 = 0

(D) x^{2} + y^{2} + 4x – 10y + 25 = 0

Answer:

(A) x^{2} + y^{2} – 4x – 10y + 25 = 0

Question 4.

The equation(s) of the tangent(s) to the circle x^{2} + y^{2} = 4 which are parallel to x + 2y + 3 = 0 are

(A) x – 2y = 2

(B) x + 2y = ±2√3

(C) x + 2y = ±2√5

(D) x – 2y = ±2√5

Answer:

(C) x + 2y = ±2√5

Question 5.

If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.

(A) \(\frac{3}{4}\)

(B) \(\frac{4}{3}\)

(C) \(\frac{1}{4}\)

(D) \(\frac{7}{4}\)

Answer:

(A) \(\frac{3}{4}\)

Hint:

Tangents are parallel to each other.

The perpendicular distance between tangents = diameter

Question 6.

The area of the circle having centre at (1, 2) and passing through (4, 6) is

(A) 5π

(B) 10π

(C) 25π

(D) 100π

Answer:

(C) 25π

Hint:

Question 7.

If a circle passes through the points (0, 0), (a, 0), and (0, b), then find the co-ordinates of its centre.

(A) \(\left(\frac{-a}{2}, \frac{-b}{2}\right)\)

(B) \(\left(\frac{a}{2}, \frac{-b}{2}\right)\)

(C) \(\left(\frac{-a}{2}, \frac{b}{2}\right)\)

(D) \(\left(\frac{a}{2}, \frac{b}{2}\right)\)

Answer:

(D) \(\left(\frac{a}{2}, \frac{b}{2}\right)\)

Question 8.

The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is

(A) x^{2} + y^{2} = 9a^{2}

(B) x^{2} + y^{2} = 16a^{2}

(C) x^{2} + y^{2} = 4a^{2}

(D) x^{2} + y^{2} = a^{2}

Answer:

(C) x^{2} + y^{2} = 4a^{2}

Hint:

Since the triangle is equilateral.

The centroid of the triangle is same as the circumcentre

and radius of the circumcircle = \(\frac{2}{3}\) (median) = \(\frac{2}{3}\)(3a) = 2a

Hence, the equation of the circumcircle whose centre is at (0, 0) and radius 2a is x^{2} + y^{2} = 4a^{2}

Question 9.

A pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at A enclosing an angle of 60. The area enclosed by these tangents and the arc of the circle is

(A) \(\frac{2}{\sqrt{3}}-\frac{\pi}{6}\)

(B) \(\sqrt{3}-\frac{\pi}{3}\)

(C) \(\frac{\pi}{3}-\frac{\sqrt{3}}{6}\)

(D) \(\sqrt{3}\left(1-\frac{\pi}{6}\right)\)

Answer:

(B) \(\sqrt{3}-\frac{\pi}{3}\)

Hint:

Question 10.

The parametric equations of the circle x^{2} + y^{2} + mx + my = 0 are

(A) x = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta\), y = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \sin \theta\)

(B) x = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta\), y = \(\frac{+m}{2}+\frac{m}{\sqrt{2}} \sin \theta\)

(C) x = 0, y = 0

(D) x = m cos θ, y = m sin θ

Answer:

(A) x = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta\), y = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \sin \theta\)

(II) Answer the following:

Question 1.

Find the centre and radius of the circle x^{2} + y^{2} – x + 2y – 3 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – x + 2y – 3 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = -1, 2f = 2 and c = -3

g = \(\frac{-1}{2}\), f = 1 and c = -3

Centre of the circle = (-g, -f) = (\(\frac{1}{2}\), -1)

and radius of the circle

Question 2.

Find the centre and radius of the circle x = 3 – 4 sin θ, y = 2 – 4 cos θ.

Solution:

Given, x = 3 – 4 sin θ, y = 2 – 4 cos θ

⇒ x – 3 = -4 sin θ, y – 2 = -4 cos θ

On squaring and adding, we get

⇒ (x – 3)^{2} + (y – 2)^{2} = (-4 sin θ)^{2} + (-4 cos θ)^{2}

⇒ (x – 3)^{2} + (y – 2)^{2} = 16 sin^{2} θ + 16 cos^{2} θ

⇒ (x – 3)^{2} + (y – 2)^{2} = 16(sin^{2} θ + cos^{2} θ)

⇒ (x – 3)^{2} + (y – 2)^{2} = 16(1)

⇒ (x – 3)^{2} + (y – 2)^{2} = 16

⇒ (x – 3)^{2} + (y – 2)^{2} = 4^{2}

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

h = 3, k = 2, r = 4

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

Question 3.

Find the equation of circle passing through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0 and whose centre is the point of intersection of lines x + y + 1 = 0 and x – 2y + 4 = 0.

Solution:

Required circle passes through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0.

x + 3y = 0

⇒ x = -3y ……..(i)

2x – 7y = 0 ……(ii)

Substituting x = -3y in (ii), we get

⇒ 2(-3y) – 7y = 0

⇒ -6y – 7y = 0

⇒ -13y = 0

⇒ y = 0

Substituting y = 0 in (i), we get

x = -3(0) = 0

Point of intersection is O(0, 0).

This point O(0, 0) lies on the circle.

Let C(h, k) be the centre of the required circle.

Since, point of intersection of lines x + y = -1 and x – 2y = -4 is the centre of circle.

∴ x = h, y = k

∴ Equations of lines become

h + k = -1 ……(iii)

h – 2k = -4 …..(iv)

By (iii) – (iv), we get

3k = 3

⇒ k = 1

Substituting k = 1 in (iii), we get

h + 1 = -1

⇒ h = -2

∴ Centre of the circle is C(-2, 1) and it passes through point O(0, 0).

Radius(r) = OC

= \(\sqrt{(0+2)^{2}+(0-1)^{2}}\)

= \(\sqrt{4+1}\)

= √5

The equation of a circle with centre at (h, k) and radius r is given by

(x – h)^{2} + (y – k)^{2} = r^{2}

Here, h = -2, k = 1

the required equation of the circle is

(x + 2)^{2} + (y – 1)^{2} = (√5)^{2}

⇒ x^{2} + 4x + 4 + y^{2} – 2y + 1 = 5

⇒ x^{2} + y^{2} + 4x – 2y = 0

Question 4.

Find the equation of the circle which passes through the origin and cuts off chords of lengths 4 and 6 on the positive side of the X-axis and Y-axis respectively.

Solution:

Let the circle cut the chord of length 4 on X-axis at point A and the chord of length 6 on the Y-axis at point B.

∴ the co-ordinates of point A are (4, 0) and co-ordinates of point B are (0, 6).

Since ∠BOA is a right angle.

AB represents the diameter of the circle.

The equation of a circle having (x_{1}, y_{1}) and (x_{2}, y_{2}) as endpoints of diameter is given by

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

Here, x_{1} = 4, y_{1} = 0, x_{2} = 0, y_{2} = 6

∴ the required equation of the circle is

⇒ (x – 4) (x – 0) + (y – 0) (y – 6) = 0

⇒ x^{2} – 4x + y^{2} – 6y = 0

⇒ x^{2} + y^{2} – 4x – 6y = 0

Question 5.

Show that the points (9, 1), (7, 9), (-2, 12) and (6, 10) are concyclic.

Solution:

Let the equation of circle passing through the points (9, 1), (7, 9), (-2, 12) be

x^{2} + y^{2} + 2gx + 2fy + c = 0 …….(i)

For point (9, 1),

Substituting x = 9 andy = 1 in (i), we get

81 + 1 + 18g + 2f + c = 0

⇒ 18g + 2f + c = -82 …..(ii)

For point (7, 9),

Substituting x = 7 andy = 9 in (i), we get

49 + 81 + 14g + 18f + c = 0

⇒ 14g + 18f + c = -130 ……(iii)

For point (-2, 12),

Substituting x = -2 and y = 12 in (i), we get

4 + 144 – 4g + 24f + c = 0

⇒ -4g + 24f + c = -148 …..(iv)

By (ii) – (iii), we get

4g – 16f = 48

⇒ g – 4f = 12 …..(v)

By (iii) – (iv), we get

18g – 6f = 18

⇒ 3g – f = 3 ……(vi)

By 3 × (v) – (vi), we get

-11f = 33

⇒ f = -3

Substituting f = -3 in (vi), we get

3g – (-3) = 3

⇒ 3g + 3 = 3

⇒ g = 0

Substituting g = 0 and f = -3 in (ii), we get

18(0) + 2(-3) + c = – 82

⇒ -6 + c = -82

⇒ c = -76

Equation of the circle becomes

x^{2} + y^{2} + 2(0)x + 2(-3)y + (-76) = 0

⇒ x^{2} + y^{2} – 6y – 76 = 0 ……(vii)

Now for the point (6, 10),

Substituting x = 6 and y = 10 in L.H.S. of (vii), we get

L.H.S = 6^{2} + 10^{2} – 6(10) – 76

= 36 + 100 – 60 – 76

= 0

= R.H.S.

∴ Point (6,10) satisfies equation (vii).

∴ the given points are concyciic.

Question 6.

The line 2x – y + 6 = 0 meets the circle x^{2} + y^{2} + 10x + 9 = 0 at A and B. Find the equation of circle with AB as diameter. Solution:

2x – y + 6 = 0

⇒ y = 2x + 6

Substituting y = 2x + 6 in x^{2} + y^{2} + 10x + 9 = 0, we get

⇒ x^{2} + (2x + 6)^{2} + 10x + 9 = 0

⇒ x^{2} + 4x^{2} + 24x + 36 + 10x + 9 = 0

⇒ 5x^{2} + 34x + 45 = 0

⇒ 5x^{2} + 25x + 9x + 45 = 0

⇒ (5x + 9) (x + 5) = 0

⇒ 5x = -9 or x = -5

⇒ x = \(\frac{-9}{5}\) or x = -5

When x = \(\frac{-9}{5}\),

y = 2 × \(\frac{-9}{5}\) + 6

= \(\frac{-18}{5}\) + 6

= \(\frac{-18+30}{5}\)

= \(\frac{12}{5}\)

∴ Point of intersection is A\(\left(\frac{-9}{5}, \frac{12}{5}\right)\)

When x = -5,

y = -10 + 6 = -4

∴ Point of intersection in B (-5, -4).

By diameter form, equation of circle with AB as diameter is

(x + \(\frac{9}{5}\)) (x + 5) + (y – \(\frac{12}{5}\)) (y + 4) = 0

⇒ (5x + 9) (x + 5) + (5y – 12) (y + 4) = 0

⇒ 5x^{2} + 25x + 9x + 45 + 5y^{2} + 20y – 12y – 48 = 0

⇒ 5x^{2} + 5y^{2} + 34x + 8y – 3 = 0

Question 7.

Show that x = -1 is a tangent to circle x^{2} + y^{2} – 4x – 2y – 4 = 0 at (-1, 1).

Solution:

Given equation of circle is x^{2} + y^{2} – 4x – 2y – 4 = 0.

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = -4, 2f = -2, c = -4

⇒ g = -2, f = -1, c = -4

The equation of a tangent to the circle

x^{2} + y^{2} + 2gx + 2fy + c = 0 at (x_{1}, y_{1}) is xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

the equation of the tangent at (-1, 1) is

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

⇒ -3x – 3 = 0

⇒ -x – 1 = 0

⇒ x = -1

∴ x = -1 is the tangent to the given circle at (-1, 1).

Question 8.

Find the equation of tangent to the circle x^{2} + y^{2} = 64 at the point P(\(\frac{2 \pi}{3}\)).

Solution:

Given equation of circle is x^{2} + y^{2} = 64

Comparing this equation with x^{2} + y^{2} = r^{2}, we get r = 8

The equation of a tangent to the circle x^{2} + y^{2} = r^{2} at P(θ) is x cos θ + y sin θ = r

∴ the equation of the tangent at P(\(\frac{2 \pi}{3}\)) is

⇒ x cos \(\frac{2 \pi}{3}\) + y sin \(\frac{2 \pi}{3}\) = 9

⇒ \(x\left(\frac{-1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=8\)

⇒ -x + √3y = 16

⇒ x – √3y + 16 = 0

Question 9.

Find the equation of locus of the point of intersection of perpendicular tangents drawn to the circle x = 5 cos θ and y = 5 sin θ.

Solution:

The locus of the point of intersection of perpendicular tangents is the director circle of the given circle.

x = 5 cos θ and y = 5 sin θ

⇒ x^{2} + y^{2} = 25 cos^{2} θ + 25 sin^{2} θ

⇒ x^{2} + y^{2} = 25 (cos^{2} θ + sin^{2} θ)

⇒ x^{2} + y^{2} = 25(1) = 25

The equation of the director circle of the circle x^{2} + y^{2} = a^{2} is x^{2} + y^{2} = 2a^{2}.

Here, a = 5

∴ the required equation is

x^{2} + y^{2} = 2(5)^{2} = 2(25)

∴ x^{2} + y^{2} = 50

Question 10.

Find the equation of the circle concentric with x^{2} + y^{2} – 4x + 6y = 1 and having radius 4 units.

Solution:

Given equation of circle is

x^{2} + y^{2} – 4x + 6y = 1 i.e., x^{2} + y^{2} – 4x + 6y – 1 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = -4, 2f = 6

⇒ g = -2, f = 3

Centre of the circle = (-g, -f) = (2, -3)

Given circle is concentric with the required circle.

∴ They have same centre.

∴ Centre of the required circle = (2, -3)

The equation of a circle with centre at (h, k) and radius r is (x – h)^{2} + (y – k)^{2} = r^{2}

Here, h = 2, k = -3 and r = 4

∴ the required equation of the circle is

(x – 2)^{2} + [y – (-3)]^{2} = 4^{2}

⇒ (x – 2)^{2} + (y + 3)^{2} = 16

⇒ x^{2} – 4x + 4 + y^{2} + 6y + 9 – 16 = 0

⇒ x^{2} + y^{2} – 4x + 6y – 3 = 0

Question 11.

Find the lengths of the intercepts made on the co-ordinate axes, by the circles.

(i) x^{2} + y^{2} – 8x + y – 20 = 0

(ii) x^{2} + y^{2} – 5x + 13y – 14 = 0

Solution:

To find x-intercept made by the circle x^{2} + y^{2} + 2gx + 2fy + c = 0,

substitute y = 0 and get a quadratic equation in x, whose roots are, say, x_{1} and x_{2}.

These values represent the abscissae of ends A and B of the x-intercept.

Length of x-intercept = |AB| = |x_{2} – x_{1}|

Similarly, substituting x = 0, we get a quadratic equation in y whose roots, say, y_{1} and y_{2} are ordinates of the ends C and D of the y-intercept.

Length of y-intercept = |CD| = |y_{2} – y_{1}|

(i) Given equation of the circle is

x^{2} + y^{2} – 8x + y – 20 = 0 ……(i)

Substituting y = 0 in (i), we get

x^{2} – 8x – 20 = 0 ……(ii)

Let AB represent the x-intercept, where

A = (x_{1}, 0), B = (x_{2}, 0)

Then from (ii),

x_{1} + x_{2} = 8 and x_{1}x_{2} = -20

(x_{1} – x_{2})^{2} = (x_{1} + x_{2})^{2} – 4x_{1}x_{2}

= (8)^{2} – 4(-20)

= 64 + 80

= 144

∴ |x_{1} – x_{2}| = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}}\) = √144 = 12

∴ Length of x – intercept =12 units

Substituting x = 0 in (i), we get

y^{2} + y – 20 = 0 …..(iii)

Let CD represent the y – intercept,

where C = (0, y_{1}) and D = (0, y_{2})

Then from (iii),

y_{1} + y_{2} = -1 and y_{1}y_{2} = -20

(y_{1} – y_{2})^{2} = (y_{1} + y_{2})^{2} – 4y_{1}y_{2}

= (-1)^{2} – 4(-20)

= 1 + 80

= 81

∴ |y_{1} – y_{2}| = \(\sqrt{\left(y_{1}-y_{2}\right)^{2}}\) = √81 = 9

∴ Length of y – intercept = 9 units.

Alternate Method:

Given equation of the circle is x^{2} + y^{2} – 8x + y – 20 = 0 ……(i)

x-intercept:

Substituting y = 0 in (i), we get

x^{2} – 8x – 20 = 0

⇒ (x – 10)(x + 2) = 0

⇒ x = 10 or x = -2

length of x-intercept = |10 – (-2)| = 12 units

y-intercept:

Substituting x = 0 in (i), we get

y^{2} + y – 20 = 0

⇒ (y + 5)(y – 4) = 0

⇒ y = -5 or y = 4

length of y-intercept = |-5 – 4| = 9 units

(ii) Given equation of the circle is

x^{2} + y^{2} – 5x + 13y – 14 = 0

Substituting y = 0 in (i), we get

x^{2} – 5x – 14 = 0 ……(ii)

Let AB represent the x-intercept, where

A = (x_{1}, 0), B = (x_{2}, 0)

Then from (ii),

x_{1} + x_{2} = 5 and x_{1}x_{2} = -14

(x_{1} – x_{2})^{2} = (x_{1} + x_{2})^{2} – 4x_{1}x_{2}

= (5)^{2} – 4(-14)

= 25 + 56

= 81

∴ |x_{1} – x_{2}| = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}}\) = √81 = 9

∴ Length of x-intercept = 9 units

Substituting x = 0 in (i), we get

y^{2} + 13y – 14 = 0 ……(iii)

Let CD represent they-intercept,

where C = (0, y_{1}), D = (0, y_{2}).

Then from (iii),

y_{1} + y_{2} = -13 and y_{1}y_{2} = -14

(y_{1} – y_{2})^{2} = (y_{1} + y_{2})^{2} – 4y_{1}y_{2}

= (-13)^{2} – 4(-14)

= 169 + 56

= 225

∴ |y_{1} – y_{2}| = \(\sqrt{\left(y_{1}-y_{2}\right)^{2}}\) = √225 = 15

∴ Length ofy-intercept = 15 units

Question 12.

Show that the circles touch each other externally. Find their point of contact and the equation of their common tangent.

(i) x^{2} + y^{2} – 4x + 10y + 20 = 0

x^{2} + y^{2} + 8x – 6y – 24 = 0

(ii) x^{2} + y^{2} – 4x – 10y + 19 = 0

x^{2} + y^{2} + 2x + 8y – 23 = 0

Solution:

(i) Given equation of the first circle is x^{2} + y^{2} – 4x + 10y + 20 = 0

Here, g = -2, f = 5, c = 20

Centre of the first circle is C_{1} = (2, -5)

Radius of the first circle is

r_{1} = \(\sqrt{(-2)^{2}+5^{2}-20}\)

= \(\sqrt{4+25-20}\)

= √9

= 3

Given equation of the second circle is x^{2} + y^{2} + 8x – 6y – 24 = 0

Here, g = 4, f = -3, c = -24

Centre of the second circle is C_{2} = (-4, 3)

Radius of the second circle is

r_{2} = \(\sqrt{4^{2}+(-3)^{2}+24}\)

= \(\sqrt{16+9+24}\)

= √49

= 7

By distance formula,

C_{1}C_{2} = \(\sqrt{(-4-2)^{2}+[3-(-5)]^{2}}\)

= \(\sqrt{36+64}\)

= √1oo

= 10

r_{1} + r_{2} = 3 + 7 = 10

Since, C_{1}C_{2} = r_{1} + r_{2}

∴ the given circles touch each other externally.

Let P(x, y) be the point of contact.

∴ P divides C_{1}C_{2} internally in the ratio r_{1} : r_{2} i.e. 3 : 7.

∴ By internal division,

Equation of common tangent is

(x^{2} + y^{2} – 4x + 10y + 20) – (x^{2} + y^{2} + 8x – 6y – 24) = 0

⇒ -4x + 10y + 20 – 8x + 6y + 24 = 0

⇒ -12x + 16y + 44 = 0

⇒ 3x – 4y – 11 = 0

(ii) Given equation of the first circle is x^{2} + y^{2} – 4x – 10y + 19 = 0

Here, g = -2, f = -5, c = 19

Centre of the first circle is C_{1} = (2, 5)

Radius of the first circle is

r_{1} = \(\sqrt{(-2)^{2}+(-5)^{2}-19}\)

= \(\sqrt{4+25-19}\)

= √10

Given equation of the second circle is x^{2} + y^{2} + 2x + 8y – 23 = 0

Here, g = 1, f = 4, c = -23

Centre of the second circle is C_{2} = (-1, -4)

Radius of the second circle is

r_{2} = \(\sqrt{(-1)^{2}+4^{2}+23}\)

= \(\sqrt{1+16+23}\)

= √40

= 2√10

By distance formula,

C_{1}C_{2} = \(\sqrt{(-1-2)^{2}+(-4-5)^{2}}\)

= \(\sqrt{9+81}\)

= √90

= 3√10

r_{1} + r_{2} = √10 + 2√10 = 3√10

Since, C_{1}C_{2} = r_{1} + r_{2}

the given circles touch each other externally.

r_{1} : r_{2} = √10 : 2√10 = 1 : 2

Let P(x, y) be the point of contact.

∴ P divides C_{1} C_{2} internally in the ratio r_{1} : r_{2} i.e. 1 : 2

∴ By internal division,

Point of contact = (1, 2)

Equation of common tangent is

(x^{2} + y^{2} – 4x – 10y + 19) – (x^{2} + y^{2} + 2x + 8y – 23) = 0

⇒ -4x – 10y + 19 – 2x – 8y + 23 = 0

⇒ -6x – 18y + 42 = 0

⇒ x + 3y – 7 = 0

Question 13.

Show that the circles touch each other internally. Find their point of contact and the equation of their common tangent.

(i) x^{2} + y^{2} – 4x – 4y – 28 = 0,

x^{2} + y^{2} – 4x – 12 = 0

(ii) x^{2} + y^{2} + 4x – 12y + 4 = 0,

x^{2} + y^{2} – 2x – 4y + 4 = 0

Solution:

(i) Given equation of the first circle is x^{2} + y^{2} – 4x – 4y – 28 = 0

Here, g = -2, f = -2, c = -28

Centre of the first circle is C_{1} = (2, 2)

Radius of the first circle is

r_{1} = \(\sqrt{(-2)^{2}+(-2)^{2}+28}\)

= \(\sqrt{4+4+28}\)

= √36

= 6

Given equation of the second circle is x^{2} + y^{2} – 4x – 12 = 0

Here, g = -2, f = 0, c = -12

Centre of the second circle is C_{2} = (2, 0)

Radius of the second circle is

r_{2} = \(\sqrt{(-2)^{2}+0^{2}+12}\)

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

= √16

= 4

By distance formula,

C_{1}C_{2} = \(\sqrt{(2-2)^{2}+(0-2)^{2}}\)

= √4

= 2

|r_{1} – r_{2}| = 6 – 4 = 2

Since, C_{1}C_{2} = |r_{1} – r_{2}|

∴ the given circles touch each other internally.

Equation of common tangent is

(x^{2} + y^{2} – 4x – 4y – 28) – (x^{2} + y^{2} – 4x – 12) = 0

⇒ -4x – 4y – 28 + 4x + 12 = 0

⇒ -4y – 16 = 0

⇒ y + 4 = 0

⇒ y = -4

Substituting y = -4 in x^{2} + y^{2} – 4x – 12 = 0, we get

⇒ x^{2} + (-4)^{2} – 4x – 12 = 0

⇒ x^{2} + 16 – 4x – 12 = 0

⇒ x^{2} – 4x + 4 = 0 .

⇒ (x – 2)^{2} = 0

⇒ x = 2

∴ Point of contact is (2, -4) and equation of common tangent is y + 4 = 0.

(ii) Given equation of the first circle is x^{2} + y^{2} + 4x – 12y + 4 = 0

Here, g = 2, f = -6, c = 4

Centre of the first circle is C_{1} = (-2, 6)

Radius of the first circle is

r_{1} = \(\sqrt{2^{2}+(-6)^{2}-4}\)

= \(\sqrt{4+36-4}\)

= √36

= 6

Given equation of the second circle is x^{2} + y^{2} – 2x – 4y + 4 = 0

Here, g = -1, f = -2, c = 4

Centre of the second circle is C_{2} = (1, 2)

Radius of the second circle is

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

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

= √1

= 1

By distance formula,

C_{1}C_{2} = \(\sqrt{[1-(-2)]^{2}+(2-6)^{2}}\)

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

= √25

= 5

|r_{1} – r_{2}| = 6 – 1 = 5

Since, C_{1}C_{2} = |r_{1} – r_{2}|

the given circles touch each other internally.

Equation of common tangent is

(x^{2} + y^{2} + 4x – 12y + 4) – (x^{2} + y^{2} – 2x – 4y + 4) = 0

⇒ 4x – 12y + 4 + 2x + 4y – 4 = 0

⇒ 6x – 8y = 0

⇒ 3x – 4y = 0

⇒ y = \(\frac{3 x}{4}\)

Substituting y = \(\frac{3 x}{4}\) in x^{2} + y^{2} – 2x – 4y + 4 = 0, we get

∴ Point of contact is \(\left(\frac{8}{5}, \frac{6}{5}\right)\) and equation of common tangent is 3x – 4y = 0.

Question 14.

Find the length of the tangent segment drawn from the point (5, 3) to the circle x^{2} + y^{2} + 10x – 6y – 17 = 0.

Solution:

Given equation of circle is x^{2} + y^{2} + 10x – 6y – 17 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = 10, 2f = -6, c = -17

⇒ g = 5, f = -3, c = -17

Centre of circle = (-g, -f) = (-5, 3)

In right angled ∆ABC,

BC^{2} = AB^{2} + AC^{2} …..[Pythagoras theorem]

⇒ (10)^{2} = AB^{2}+ (√51)^{2}

⇒ AB^{2} = 100 – 51 = √49

⇒ AB = 7

∴ Length of the tangent segment from (5, 3) is 7 units.

Alternate method:

Given equation of circle is x^{2} + y^{2} + 10x – 6y – 17 = 0

Here, g = 5, f = -3, c = -17

Length of the tangent segment to the circle x^{2} + y^{2} + 2gx + 2fy + c = 0 from the point (x_{1}, y_{1}) is \(\sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c}\)

Length of the tangent segment from (5, 3)

= \(\sqrt{(5)^{2}+(3)^{2}+10(5)-6(3)-17}\)

= \(\sqrt{25+9+50-18-17}\)

= √49

= 7 units

Question 15.

Find the value of k, if the length of the tangent segment from the point (8, -3) to the circle x^{2} + y^{2} – 2x + ky – 23 = 0 is √10.

Solution:

Given equation of the circle is x^{2} + y^{2} – 2x + ky – 23 = 0

Here, g = -1, f = \(\frac{\mathrm{k}}{2}\), c = -23

Length of the tangent segment to the circle x^{2} + y^{2} + 2gx + 2fy + c = 0 from the point (x_{1}, y_{1}) is \(\sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c}\)

Length of the tangent segment from (8, -3) = √10

⇒ \(\sqrt{8^{2}+(-3)^{2}-2(8)+k(-3)-23}=\sqrt{10}\)

⇒ 64 + 9 – 16 – 3k – 23 = 10 …..[Squaring both the sides]

⇒ 34 – 3k = 10

⇒ 3k = 24

⇒ k = 8

Question 16.

Find the equation of tangent to circle x^{2} + y^{2} – 6x – 4y = 0, at the point (6, 4) on it.

Solution:

Given equation of the circle is x^{2} + y^{2} – 6x – 4y = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = -6, 2f = -4, c = 0

⇒ g = -3, f = -2, c = 0

The equation of a tangent to the circle x^{2} + y^{2} + 2gx + 2fy + c = 0 at (x_{1}, y_{1}) is

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

the equation of the tangent at (6, 4) is

x(6) + y(4) – 3(x + 6) – 2(y + 4) + 0 = 0

⇒ 6x + 4y – 3x – 18 – 2y – 8 = 0

⇒ 3x + 2y – 26 = 0

Alternate method:

Given equation of the circle is x^{2} + y^{2} – 6x – 4y = 0

x(x – 6) + y(y – 4) = 0, which is in diameter form where (0, 0) and (6, 4) are endpoints of diameter.

Slope of OP = \(\frac{4-0}{6-0}=\frac{2}{3}\)

Since, OP is perpendicular to the required tangent.

Slope of the required tangent = \(\frac{-3}{2}\)

the equation of the tangent at (6, 4) is

y – 4 = \(\frac{-3}{2}\) (x – 6)

⇒ 2(y – 4) = 3(x – 6)

⇒ 2y – 8 = -3x + 18

⇒ 3x + 2y – 26 = 0

Question 17.

Fihd the equation of tangent to circle x^{2} + y^{2} = 5, at the point (1, -2) on it.

Solution:

Given equation of the circle is x^{2} + y^{2} = 5

Comparing this equation with x^{2} + y^{2} = r^{2}, we get

r^{2} = 5

The equation of a tangent to the circle x^{2} + y^{2} = r^{2} at (x_{1}, y_{1}) is xx_{1} + yy_{1} = r^{2}

the equation of the tangent at (1, -2) is

x(1) + y(-2) = 5

⇒ x – 2y = 5

Question 18.

Find the equation of tangent to circle x = 5 cos θ, y = 5 sin θ, at the point θ = \(\frac{\pi}{3}\) on it.

Solution:

The equation of a tangent to the circle x^{2} + y^{2} = r^{2} at P(θ) is x cos θ + y sin θ = r

Here, r = 5, θ = \(\frac{\pi}{3}\)

the equation of the tangent at P(\(\frac{\pi}{3}\)) is

x cos \(\frac{\pi}{3}\) + y sin \(\frac{\pi}{3}\) = 5

⇒ \(x\left(\frac{1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=5\)

⇒ x + √3y = 10

Question 19.

Show that 2x + y + 6 = 0 is a tangent to x^{2} + y^{2} + 2x – 2y – 3 = 0. Find its point of contact.

Solution:

Given equation of circle is

x^{2} + y^{2} + 2x – 2y – 3 = 0 ….(i)

Given equation of line is 2x + y + 6 = 0

y = -6 – 2x ……(ii)

Substituting y = -6 – 2x in (i), we get

x + (-6 – 2x)^{2} + 2x – 2(-6 – 2x) – 3 = 0

⇒ x^{2} + 36 + 24x + 4x^{2} + 2x + 12 + 4x – 3 = 0

⇒ 5x^{2} + 30x + 45 = 0

⇒ x^{2} + 6x + 9 = 0

⇒ (x + 3)^{2} = 0

⇒ x = -3

Since, the roots are equal.

∴ 2x + y + 6 = 0 is a tangent to x^{2} + y^{2} + 2x – 2y – 3 = 0

Substituting x = -3 in (ii), we get

y = -6 – 2(-3) = -6 + 6 = 0

Point of contact = (-3, 0)

Question 20.

If the tangent at (3, -4) to the circle x^{2} + y^{2} = 25 touches the circle x^{2} + y^{2} + 8x – 4y + c = 0, find c.

Solution:

The equation of a tangent to the circle

x^{2} + y^{2} = r^{2} at (x_{1}, y_{1}) is xx_{1} + yy_{1} = r^{2}

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

x(3) + y(-4) = 25

⇒ 3x – 4y – 25 = 0 ……(i)

Given equation of circle is x^{2} + y^{2} + 8x – 4y + c = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = 8, 2f = -4

⇒ g = 4, f = -2

∴ C = (-4, 2) and r = \(\sqrt{4^{2}+(-2)^{2}-c}=\sqrt{20-c}\)

Since line (i) is a tangent to this circle also, the perpendicular distance from C(-4, 2) to line (i) is equal to radius r.

Question 21.

Find the equations of the tangents to the circle x^{2} + y^{2} = 16 with slope -2.

Solution:

Given equation of the circle is x^{2} + y^{2} = 16

Comparing this equation with x^{2} + y^{2} = a^{2}, we get

a^{2} = 16

Equations of the tangents to the circle x^{2} + y^{2} = a^{2} with slope m are

\(y=m x \pm \sqrt{a^{2}\left(1+m^{2}\right)}\)

Here, m = -2, a^{2} = 16

the required equations of the tangents are

y = \(-2 x \pm \sqrt{16\left[1+(-2)^{2}\right]}\)

⇒ y = \(-2 x \pm \sqrt{16(5)}\)

⇒ y = -2x ± 4√5

⇒ 2x + y ± 4√5 = 0

Question 22.

Find the equations of the tangents to the circle x^{2} + y^{2} = 4 which are parallel to 3x + 2y + 1 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} = 4

Comparing this equation with x^{2} + y^{2} = a^{2}, we get

a^{2} = 4

Given equation of the line is 3x + 2y + 1 = 0

Slope of this line = \(\frac{-3}{2}\)

Since, the required tangents are parallel to the given line.

Slope of required tangents (m) = \(\frac{-3}{2}\)

Equations of the tangents to the circle x^{2} + y^{2} = a^{2} with slope m are

y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)

the required equations of the tangents are

Question 23.

Find the equations of the tangents to the circle x^{2} + y^{2} = 36 which are perpendicular to the line 5x + y = 2.

Solution:

Given equation of the circle is x^{2} + y^{2} = 36

Comparing this equaiton with x^{2} + y^{2} = a^{2}, we get

a^{2} = 36

Given equation of line is 5x + y = 2

Slope of this line = -5

Since, the required tangents are perpendicular to the given line.

Slope of required tangents (m) = \(\frac{1}{5}\)

Equations of the tangents to the circle x^{2} + y^{2} = a^{2} with slope m are

y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)

the required equations of the tangents are

Question 24.

Find the equations of the tangents to the circle x^{2} + y^{2} – 2x + 8y – 23 = 0 having slope 3.

Solution:

Let the equation of the tangent with slope 3 be y = 3x + c.

3x – y + c = 0 ……(i)

Given equation of circle is x^{2} + y^{2} – 2x + 8y – 23 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = -2, 2f = 8, c = -23

g = -1, f = 4, c = -23

The centre of the circle is C(1, -4)

and its radius = \(\sqrt{1+16+23}\)

= √40

= 2√10

Since line (i) is a tangent to this circle the perpendicular distance from C(1, -4) to line (i) is equal to radius r.

\(\left|\frac{3(1)+4+c}{\sqrt{9+1}}\right|\) = 2√10

⇒ \(\left|\frac{7+c}{\sqrt{10}}\right|\) = 2√10

⇒ (7 + c) = ± 20

⇒ 7 + c = 20 or 7 + c = -20

⇒ c = 13 or c = – 27

∴ Equations of the tangents are 3x – y + 13 = 0 and 3x – y – 21 = 0

Question 25.

Find the equation of the locus of a point, the tangents from which to the circle x^{2} + y^{2} = 9 are at right angles.

Solution:

Given equation of the circle is x^{2} + y^{2} = 9

Comparing this equation with x^{2} + y^{2} = a^{2}, we get

a^{2} = 9

The locus of the point of intersection of perpendicular tangents is the director circle of the given circle.

The equation of the director circle of the circle x^{2} + y^{2} = a^{2} is x^{2} + y^{2} = 2a^{2}.

the required equation is

x^{2} + y^{2} = 2(9)

x^{2} + y^{2} = 18

Alternate method:

Given equation of the circle is x^{2} + y^{2} = 9

Comparing this equation with x^{2} + y^{2} = a^{2}, we get a^{2} = 9

Let P(x_{1}, y_{1}) be a point on the required locus.

Equations of the tangents to the circle x^{2} + y^{2} = a^{2} with slope m are

y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)

∴ Equations of the tangents are

y = mx ± \(\sqrt{9\left(\mathrm{~m}^{2}+1\right)}\)

⇒ y = mx ± 3\(\sqrt{1+m^{2}}\)

Since, these tangents pass through (x_{1}, y_{1}).

y_{1} = mx_{1} ± 3\(\sqrt{1+m^{2}}\)

⇒ y_{1} – mx_{1} = ± 3\(\sqrt{1+m^{2}}\)

⇒ (y_{1} – mx_{1})^{2} = 9(1 + m^{2}) ……[Squaring both the sides]

⇒ \(y_{1}^{2}-2 m x_{1} y_{1}+m^{2} x_{1}^{2}=9+9 m^{2}\)

⇒ \(\left(x_{1}^{2}-9\right) \mathrm{m}^{2}-2 \mathrm{~m} x_{1} y_{1}+\left(y_{1}^{2}-9\right)=0\)

This is a quadratic equation which has two roots m_{1} and m_{2}.

m_{1}m_{2} = \(\frac{y_{1}^{2}-9}{x_{1}^{2}-9}\)

Since, the tangents are at right angles.

m_{1}m_{2} = -1

⇒ \(\frac{y_{1}^{2}-9}{x_{1}^{2}-9}=-1\)

⇒ \(y_{1}^{2}-9=9-x_{1}^{2}\)

⇒ \(x_{1}^{2}+y_{1}^{2}=18\)

Equation of the locus of point P is x^{2} + y^{2} = 18.

Question 26.

Tangents to the circle x^{2} + y^{2} = a^{2} with inclinations, θ_{1} and θ_{2} intersect in P. Find the locus of P such that

(i) tan θ_{1} + tan θ_{2} = 0

(ii) cot θ_{1} + cot θ_{2} = 5

(iii) cot θ_{1} . cot θ_{2} = c

Solution:

Let P(x_{1}, y_{1}) be a point on the required locus.

Equations of the tangents to the circle x^{2} + y^{2} = a^{2} with slope m are

y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)

Since, these tangents pass through (x_{1}, y_{1}).