Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Miscellaneous Exercise 6
(I) Choose the correct alternative.
Question 1
Maharashtra Board Solution
Equation of a circle which passes through (3, 6) and touches the axes is (A) x2 + y2 + 6x + 6y + 3 = 0 (B) x2 + y2 – 6x – 6y – 9 = 0 (C) x2 + y2 – 6x – 6y + 9 = 0 (D) x2 + y2 – 6x + 6y – 3 = 0
Solution & Step-by-Step Answer:
(C) x2 + y2 – 6x – 6y + 9 = 0
Question 2
Maharashtra Board Solution
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) x2 + y2 – 2x + 2y = 40 (B) x2 + y2 – 2x – 2y = 47 (C) x2 + y2 – 2x + 2y = 47 (D) x2 + y2 – 2x – 2y = 40
Solution & Step-by-Step Answer:
(C) x2 + y2 – 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 πr2 = 154 = 154 r2 = 154 × = 49 ∴ r = 7 equation of the circle is (x – 1)2 + (y + 1)2 = 72 x2 + y2 – 2x + 2y = 47
Question 3
Maharashtra Board Solution
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) x2 + y2 – 4x – 10y + 25 = 0 (B) x2 + y2 – 4x – 10y – 25 = 0 (C) x2 + y2 – 4x + 10y – 25 = 0 (D) x2 + y2 + 4x – 10y + 25 = 0
Solution & Step-by-Step Answer:
(A) x2 + y2 – 4x – 10y + 25 = 0
Question 4
Maharashtra Board Solution
The equation(s) of the tangent(s) to the circle x2 + y2 = 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
Solution & Step-by-Step Answer:
(C) x + 2y = ±2√5
Question 5
Maharashtra Board Solution
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) (B) (C) (D)
Solution & Step-by-Step Answer:
(A) Hint: Tangents are parallel to each other. The perpendicular distance between tangents = diameter
Question 6
Maharashtra Board Solution
The area of the circle having centre at (1, 2) and passing through (4, 6) is (A) 5π (B) 10π (C) 25π (D) 100π
Solution & Step-by-Step Answer:
(C) 25π Hint:
Question 7
Maharashtra Board Solution
If a circle passes through the points (0, 0), (a, 0), and (0, b), then find the co-ordinates of its centre. (A) (B) (C) (D)
Solution & Step-by-Step Answer:
(D)
Question 8
Maharashtra Board Solution
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) x2 + y2 = 9a2 (B) x2 + y2 = 16a2 (C) x2 + y2 = 4a2 (D) x2 + y2 = a2
Solution & Step-by-Step Answer:
(C) x2 + y2 = 4a2 Hint: Since the triangle is equilateral. The centroid of the triangle is same as the circumcentre and radius of the circumcircle = (median) = (3a) = 2a Hence, the equation of the circumcircle whose centre is at (0, 0) and radius 2a is x2 + y2 = 4a2
Question 9
Maharashtra Board Solution
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) (B) (C) (D)
Solution & Step-by-Step Answer:
(B) Hint:
Question 10
Maharashtra Board Solution
The parametric equations of the circle x2 + y2 + mx + my = 0 are (A) x = , y = (B) x = , y = (C) x = 0, y = 0 (D) x = m cos θ, y = m sin θ
Solution & Step-by-Step Answer:
(A) x = , y =
(II) Answer the following:
Question 1
Maharashtra Board Solution
Find the centre and radius of the circle x2 + y2 – x + 2y – 3 = 0.
Solution & Step-by-Step Answer:
Given equation of the circle is x2 + y2 – x + 2y – 3 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = -1, 2f = 2 and c = -3 g = , f = 1 and c = -3 Centre of the circle = (-g, -f) = (, -1) and radius of the circle
Question 2
Maharashtra Board Solution
Find the centre and radius of the circle x = 3 – 4 sin θ, y = 2 – 4 cos θ.
Solution & Step-by-Step Answer:
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 sin2 θ + 16 cos2 θ ⇒ (x – 3)2 + (y – 2)2 = 16(sin2 θ + cos2 θ) ⇒ (x – 3)2 + (y – 2)2 = 16(1) ⇒ (x – 3)2 + (y – 2)2 = 16 ⇒ (x – 3)2 + (y – 2)2 = 42 Comparing this equation with (x – h)2 + (y – k)2 = r2, we get h = 3, k = 2, r = 4 ∴ Centre of the circle is (3, 2) and radius is 4.
Question 3
Maharashtra Board Solution
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 & Step-by-Step Answer:
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 = = = √5 The equation of a circle with centre at (h, k) and radius r is given by (x – h)2 + (y – k)2 = r2 Here, h = -2, k = 1 the required equation of the circle is (x + 2)2 + (y – 1)2 = (√5)2 ⇒ x2 + 4x + 4 + y2 – 2y + 1 = 5 ⇒ x2 + y2 + 4x – 2y = 0
Question 4
Maharashtra Board Solution
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 & Step-by-Step Answer:
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 (x1, y1) and (x2, y2) as endpoints of diameter is given by (x – x1) (x – x2) + (y – y1) (y – y2) = 0 Here, x1 = 4, y1 = 0, x2 = 0, y2 = 6 ∴ the required equation of the circle is ⇒ (x – 4) (x – 0) + (y – 0) (y – 6) = 0 ⇒ x2 – 4x + y2 – 6y = 0 ⇒ x2 + y2 – 4x – 6y = 0
Question 5
Maharashtra Board Solution
Show that the points (9, 1), (7, 9), (-2, 12) and (6, 10) are concyclic.
Solution & Step-by-Step Answer:
Let the equation of circle passing through the points (9, 1), (7, 9), (-2, 12) be x2 + y2 + 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 x2 + y2 + 2(0)x + 2(-3)y + (-76) = 0 ⇒ x2 + y2 – 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 = 62 + 102 – 6(10) – 76 = 36 + 100 – 60 – 76 = 0 = R.H.S. ∴ Point (6,10) satisfies equation (vii). ∴ the given points are concyciic.
Question 6
Maharashtra Board Solution
The line 2x – y + 6 = 0 meets the circle x2 + y2 + 10x + 9 = 0 at A and B. Find the equation of circle with AB as diameter.
Solution & Step-by-Step Answer:
2x – y + 6 = 0 ⇒ y = 2x + 6 Substituting y = 2x + 6 in x2 + y2 + 10x + 9 = 0, we get ⇒ x2 + (2x + 6)2 + 10x + 9 = 0 ⇒ x2 + 4x2 + 24x + 36 + 10x + 9 = 0 ⇒ 5x2 + 34x + 45 = 0 ⇒ 5x2 + 25x + 9x + 45 = 0 ⇒ (5x + 9) (x + 5) = 0 ⇒ 5x = -9 or x = -5 ⇒ x = or x = -5 When x = , y = 2 × + 6 = + 6 = = ∴ Point of intersection is A 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 + ) (x + 5) + (y – ) (y + 4) = 0 ⇒ (5x + 9) (x + 5) + (5y – 12) (y + 4) = 0 ⇒ 5x2 + 25x + 9x + 45 + 5y2 + 20y – 12y – 48 = 0 ⇒ 5x2 + 5y2 + 34x + 8y – 3 = 0
Question 7
Maharashtra Board Solution
Show that x = -1 is a tangent to circle x2 + y2 – 4x – 2y – 4 = 0 at (-1, 1).
Solution & Step-by-Step Answer:
Given equation of circle is x2 + y2 – 4x – 2y – 4 = 0. Comparing this equation with x2 + y2 + 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 x2 + y2 + 2gx + 2fy + c = 0 at (x1, y1) is xx1 + yy1 + g(x + x1) + f(y + y1) + 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
Maharashtra Board Solution
Find the equation of tangent to the circle x2 + y2 = 64 at the point P().
Solution & Step-by-Step Answer:
Given equation of circle is x2 + y2 = 64 Comparing this equation with x2 + y2 = r2, we get r = 8 The equation of a tangent to the circle x2 + y2 = r2 at P(θ) is x cos θ + y sin θ = r ∴ the equation of the tangent at P() is ⇒ x cos + y sin = 9 ⇒ ⇒ -x + √3y = 16 ⇒ x – √3y + 16 = 0
Question 9
Maharashtra Board Solution
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 & Step-by-Step Answer:
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 ⇒ x2 + y2 = 25 cos2 θ + 25 sin2 θ ⇒ x2 + y2 = 25 (cos2 θ + sin2 θ) ⇒ x2 + y2 = 25(1) = 25 The equation of the director circle of the circle x2 + y2 = a2 is x2 + y2 = 2a2. Here, a = 5 ∴ the required equation is x2 + y2 = 2(5)2 = 2(25) ∴ x2 + y2 = 50
Question 10
Maharashtra Board Solution
Find the equation of the circle concentric with x2 + y2 – 4x + 6y = 1 and having radius 4 units.
Solution & Step-by-Step Answer:
Given equation of circle is x2 + y2 – 4x + 6y = 1 i.e., x2 + y2 – 4x + 6y – 1 = 0 Comparing this equation with x2 + y2 + 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 = r2 Here, h = 2, k = -3 and r = 4 ∴ the required equation of the circle is (x – 2)2 + [y – (-3)]2 = 42 ⇒ (x – 2)2 + (y + 3)2 = 16 ⇒ x2 – 4x + 4 + y2 + 6y + 9 – 16 = 0 ⇒ x2 + y2 – 4x + 6y – 3 = 0
Question 11
Maharashtra Board Solution
Find the lengths of the intercepts made on the co-ordinate axes, by the circles. (i) x2 + y2 – 8x + y – 20 = 0 (ii) x2 + y2 – 5x + 13y – 14 = 0
Solution & Step-by-Step Answer:
To find x-intercept made by the circle x2 + y2 + 2gx + 2fy + c = 0, substitute y = 0 and get a quadratic equation in x, whose roots are, say, x1 and x2. These values represent the abscissae of ends A and B of the x-intercept. Length of x-intercept = |AB| = |x2 – x1| Similarly, substituting x = 0, we get a quadratic equation in y whose roots, say, y1 and y2 are ordinates of the ends C and D of the y-intercept. Length of y-intercept = |CD| = |y2 – y1| (i) Given equation of the circle is x2 + y2 – 8x + y – 20 = 0 ……(i) Substituting y = 0 in (i), we get x2 – 8x – 20 = 0 ……(ii) Let AB represent the x-intercept, where A = (x1, 0), B = (x2, 0) Then from (ii), x1 + x2 = 8 and x1x2 = -20 (x1 – x2)2 = (x1 + x2)2 – 4x1x2 = (8)2 – 4(-20) = 64 + 80 = 144 ∴ |x1 – x2| = = √144 = 12 ∴ Length of x – intercept =12 units Substituting x = 0 in (i), we get y2 + y – 20 = 0 …..(iii) Let CD represent the y – intercept, where C = (0, y1) and D = (0, y2) Then from (iii), y1 + y2 = -1 and y1y2 = -20 (y1 – y2)2 = (y1 + y2)2 – 4y1y2 = (-1)2 – 4(-20) = 1 + 80 = 81 ∴ |y1 – y2| = = √81 = 9 ∴ Length of y – intercept = 9 units.
Alternate Method:
Given equation of the circle is x2+ y2– 8x + y – 20 = 0 ……(i)
x-intercept:
Substituting y = 0 in (i), we get
x2– 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
y2+ 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
x2+ y2– 5x + 13y – 14 = 0
Substituting y = 0 in (i), we get
x2– 5x – 14 = 0 ……(ii)
Let AB represent the x-intercept, where
A = (x1, 0), B = (x2, 0)
Then from (ii),
x1+ x2= 5 and x1x2= -14
(x1– x2)2= (x1+ x2)2– 4x1x2
= (5)2– 4(-14)
= 25 + 56
= 81
∴ |x1– x2| = = √81 = 9
∴ Length of x-intercept = 9 units
Substituting x = 0 in (i), we get
y2+ 13y – 14 = 0 ……(iii)
Let CD represent they-intercept,
where C = (0, y1), D = (0, y2).
Then from (iii),
y1+ y2= -13 and y1y2= -14
(y1– y2)2= (y1+ y2)2– 4y1y2
= (-13)2– 4(-14)
= 169 + 56
= 225
∴ |y1– y2| = = √225 = 15
∴ Length ofy-intercept = 15 units
Question 12
Maharashtra Board Solution
Show that the circles touch each other externally. Find their point of contact and the equation of their common tangent. (i) x2 + y2 – 4x + 10y + 20 = 0 x2 + y2 + 8x – 6y – 24 = 0 (ii) x2 + y2 – 4x – 10y + 19 = 0 x2 + y2 + 2x + 8y – 23 = 0
Solution & Step-by-Step Answer:
(i) Given equation of the first circle is x2 + y2 – 4x + 10y + 20 = 0 Here, g = -2, f = 5, c = 20 Centre of the first circle is C1 = (2, -5) Radius of the first circle is r1 = = = √9 = 3 Given equation of the second circle is x2 + y2 + 8x – 6y – 24 = 0 Here, g = 4, f = -3, c = -24 Centre of the second circle is C2 = (-4, 3) Radius of the second circle is r2 = = = √49 = 7 By distance formula, C1C2 = = = √1oo = 10 r1 + r2 = 3 + 7 = 10 Since, C1C2 = r1 + r2 ∴ the given circles touch each other externally. Let P(x, y) be the point of contact. ∴ P divides C1C2 internally in the ratio r1 : r2 i.e. 3 : 7. ∴ By internal division, Equation of common tangent is (x2 + y2 – 4x + 10y + 20) – (x2 + y2 + 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 x2+ y2– 4x – 10y + 19 = 0
Here, g = -2, f = -5, c = 19
Centre of the first circle is C1= (2, 5)
Radius of the first circle is
r1=
=
= √10
Given equation of the second circle is x2+ y2+ 2x + 8y – 23 = 0
Here, g = 1, f = 4, c = -23
Centre of the second circle is C2= (-1, -4)
Radius of the second circle is
r2=
=
= √40
= 2√10
By distance formula,
C1C2=
=
= √90
= 3√10
r1+ r2= √10 + 2√10 = 3√10
Since, C1C2= r1+ r2
the given circles touch each other externally.
r1: r2= √10 : 2√10 = 1 : 2
Let P(x, y) be the point of contact.
∴ P divides C1C2internally in the ratio r1: r2i.e. 1 : 2
∴ By internal division,
Point of contact = (1, 2)
Equation of common tangent is
(x2+ y2– 4x – 10y + 19) – (x2+ y2+ 2x + 8y – 23) = 0
⇒ -4x – 10y + 19 – 2x – 8y + 23 = 0
⇒ -6x – 18y + 42 = 0
⇒ x + 3y – 7 = 0
Question 13
Maharashtra Board Solution
Show that the circles touch each other internally. Find their point of contact and the equation of their common tangent. (i) x2 + y2 – 4x – 4y – 28 = 0, x2 + y2 – 4x – 12 = 0 (ii) x2 + y2 + 4x – 12y + 4 = 0, x2 + y2 – 2x – 4y + 4 = 0
Solution & Step-by-Step Answer:
(i) Given equation of the first circle is x2 + y2 – 4x – 4y – 28 = 0 Here, g = -2, f = -2, c = -28 Centre of the first circle is C1 = (2, 2) Radius of the first circle is r1 = = = √36 = 6 Given equation of the second circle is x2 + y2 – 4x – 12 = 0 Here, g = -2, f = 0, c = -12 Centre of the second circle is C2 = (2, 0) Radius of the second circle is r2 = = = √16 = 4 By distance formula, C1C2 = = √4 = 2 |r1 – r2| = 6 – 4 = 2 Since, C1C2 = |r1 – r2| ∴ the given circles touch each other internally. Equation of common tangent is (x2 + y2 – 4x – 4y – 28) – (x2 + y2 – 4x – 12) = 0 ⇒ -4x – 4y – 28 + 4x + 12 = 0 ⇒ -4y – 16 = 0 ⇒ y + 4 = 0 ⇒ y = -4 Substituting y = -4 in x2 + y2 – 4x – 12 = 0, we get ⇒ x2 + (-4)2 – 4x – 12 = 0 ⇒ x2 + 16 – 4x – 12 = 0 ⇒ x2 – 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 x2+ y2+ 4x – 12y + 4 = 0
Here, g = 2, f = -6, c = 4
Centre of the first circle is C1= (-2, 6)
Radius of the first circle is
r1=
=
= √36
= 6
Given equation of the second circle is x2+ y2– 2x – 4y + 4 = 0
Here, g = -1, f = -2, c = 4
Centre of the second circle is C2= (1, 2)
Radius of the second circle is
r2=
=
= √1
= 1
By distance formula,
C1C2=
=
= √25
= 5
|r1– r2| = 6 – 1 = 5
Since, C1C2= |r1– r2|
the given circles touch each other internally.
Equation of common tangent is
(x2+ y2+ 4x – 12y + 4) – (x2+ y2– 2x – 4y + 4) = 0
⇒ 4x – 12y + 4 + 2x + 4y – 4 = 0
⇒ 6x – 8y = 0
⇒ 3x – 4y = 0
⇒ y =
Substituting y = in x2+ y2– 2x – 4y + 4 = 0, we get
∴ Point of contact is and equation of common tangent is 3x – 4y = 0.
Question 14
Maharashtra Board Solution
Find the length of the tangent segment drawn from the point (5, 3) to the circle x2 + y2 + 10x – 6y – 17 = 0.
Solution & Step-by-Step Answer:
Given equation of circle is x2 + y2 + 10x – 6y – 17 = 0 Comparing this equation with x2 + y2 + 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, BC2 = AB2 + AC2 …..[Pythagoras theorem] ⇒ (10)2 = AB2+ (√51)2 ⇒ AB2 = 100 – 51 = √49 ⇒ AB = 7 ∴ Length of the tangent segment from (5, 3) is 7 units.
Alternate method:
Given equation of circle is x2+ y2+ 10x – 6y – 17 = 0
Here, g = 5, f = -3, c = -17
Length of the tangent segment to the circle x2+ y2+ 2gx + 2fy + c = 0 from the point (x1, y1) is
Length of the tangent segment from (5, 3)
=
=
= √49
= 7 units
Question 15
Maharashtra Board Solution
Find the value of k, if the length of the tangent segment from the point (8, -3) to the circle x2 + y2 – 2x + ky – 23 = 0 is √10.
Solution & Step-by-Step Answer:
Given equation of the circle is x2 + y2 – 2x + ky – 23 = 0 Here, g = -1, f = , c = -23 Length of the tangent segment to the circle x2 + y2 + 2gx + 2fy + c = 0 from the point (x1, y1) is Length of the tangent segment from (8, -3) = √10 ⇒ ⇒ 64 + 9 – 16 – 3k – 23 = 10 …..[Squaring both the sides] ⇒ 34 – 3k = 10 ⇒ 3k = 24 ⇒ k = 8
Question 16
Maharashtra Board Solution
Find the equation of tangent to circle x2 + y2 – 6x – 4y = 0, at the point (6, 4) on it.
Solution & Step-by-Step Answer:
Given equation of the circle is x2 + y2 – 6x – 4y = 0 Comparing this equation with x2 + y2 + 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 x2 + y2 + 2gx + 2fy + c = 0 at (x1, y1) is xx1 + yy1 + g(x + x1) + f(y + y1) + 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 x2+ y2– 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 =
Since, OP is perpendicular to the required tangent.
Slope of the required tangent =
the equation of the tangent at (6, 4) is
y – 4 = (x – 6)
⇒ 2(y – 4) = 3(x – 6)
⇒ 2y – 8 = -3x + 18
⇒ 3x + 2y – 26 = 0
Question 17
Maharashtra Board Solution
Fihd the equation of tangent to circle x2 + y2 = 5, at the point (1, -2) on it.
Solution & Step-by-Step Answer:
Given equation of the circle is x2 + y2 = 5 Comparing this equation with x2 + y2 = r2, we get r2 = 5 The equation of a tangent to the circle x2 + y2 = r2 at (x1, y1) is xx1 + yy1 = r2 the equation of the tangent at (1, -2) is x(1) + y(-2) = 5 ⇒ x – 2y = 5
Question 18
Maharashtra Board Solution
Find the equation of tangent to circle x = 5 cos θ, y = 5 sin θ, at the point θ = on it.
Solution & Step-by-Step Answer:
The equation of a tangent to the circle x2 + y2 = r2 at P(θ) is x cos θ + y sin θ = r Here, r = 5, θ = the equation of the tangent at P() is x cos + y sin = 5 ⇒ ⇒ x + √3y = 10
Question 19
Maharashtra Board Solution
Show that 2x + y + 6 = 0 is a tangent to x2 + y2 + 2x – 2y – 3 = 0. Find its point of contact.
Solution & Step-by-Step Answer:
Given equation of circle is x2 + y2 + 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 ⇒ x2 + 36 + 24x + 4x2 + 2x + 12 + 4x – 3 = 0 ⇒ 5x2 + 30x + 45 = 0 ⇒ x2 + 6x + 9 = 0 ⇒ (x + 3)2 = 0 ⇒ x = -3 Since, the roots are equal. ∴ 2x + y + 6 = 0 is a tangent to x2 + y2 + 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
Maharashtra Board Solution
If the tangent at (3, -4) to the circle x2 + y2 = 25 touches the circle x2 + y2 + 8x – 4y + c = 0, find c.
Solution & Step-by-Step Answer:
The equation of a tangent to the circle x2 + y2 = r2 at (x1, y1) is xx1 + yy1 = r2 Equation of the tangent at (3, -4) is x(3) + y(-4) = 25 ⇒ 3x – 4y – 25 = 0 ……(i) Given equation of circle is x2 + y2 + 8x – 4y + c = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 8, 2f = -4 ⇒ g = 4, f = -2 ∴ C = (-4, 2) and r = 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
Maharashtra Board Solution
Find the equations of the tangents to the circle x2 + y2 = 16 with slope -2.
Solution & Step-by-Step Answer:
Given equation of the circle is x2 + y2 = 16 Comparing this equation with x2 + y2 = a2, we get a2 = 16 Equations of the tangents to the circle x2 + y2 = a2 with slope m are Here, m = -2, a2 = 16 the required equations of the tangents are y = ⇒ y = ⇒ y = -2x ± 4√5 ⇒ 2x + y ± 4√5 = 0
Question 22
Maharashtra Board Solution
Find the equations of the tangents to the circle x2 + y2 = 4 which are parallel to 3x + 2y + 1 = 0.
Solution & Step-by-Step Answer:
Given equation of the circle is x2 + y2 = 4 Comparing this equation with x2 + y2 = a2, we get a2 = 4 Given equation of the line is 3x + 2y + 1 = 0 Slope of this line = Since, the required tangents are parallel to the given line. Slope of required tangents (m) = Equations of the tangents to the circle x2 + y2 = a2 with slope m are y = mx ± the required equations of the tangents are
Question 23
Maharashtra Board Solution
Find the equations of the tangents to the circle x2 + y2 = 36 which are perpendicular to the line 5x + y = 2.
Solution & Step-by-Step Answer:
Given equation of the circle is x2 + y2 = 36 Comparing this equaiton with x2 + y2 = a2, we get a2 = 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) = Equations of the tangents to the circle x2 + y2 = a2 with slope m are y = mx ± the required equations of the tangents are
Question 24
Maharashtra Board Solution
Find the equations of the tangents to the circle x2 + y2 – 2x + 8y – 23 = 0 having slope 3.
Solution & Step-by-Step Answer:
Let the equation of the tangent with slope 3 be y = 3x + c. 3x – y + c = 0 ……(i) Given equation of circle is x2 + y2 – 2x + 8y – 23 = 0 Comparing this equation with x2 + y2 + 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 = = √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. = 2√10 ⇒ = 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
Maharashtra Board Solution
Find the equation of the locus of a point, the tangents from which to the circle x2 + y2 = 9 are at right angles.
Solution & Step-by-Step Answer:
Given equation of the circle is x2 + y2 = 9 Comparing this equation with x2 + y2 = a2, we get a2 = 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 x2 + y2 = a2 is x2 + y2 = 2a2. the required equation is x2 + y2 = 2(9) x2 + y2 = 18
Alternate method:
Given equation of the circle is x2+ y2= 9
Comparing this equation with x2+ y2= a2, we get a2= 9
Let P(x1, y1) be a point on the required locus.
Equations of the tangents to the circle x2+ y2= a2with slope m are
y = mx ±
∴ Equations of the tangents are
y = mx ±
⇒ y = mx ± 3
Since, these tangents pass through (x1, y1).
y1= mx1± 3
⇒ y1– mx1= ± 3
⇒ (y1– mx1)2= 9(1 + m2) ……[Squaring both the sides]
⇒
⇒
This is a quadratic equation which has two roots m1and m2.
m1m2=
Since, the tangents are at right angles.
m1m2= -1
⇒
⇒
⇒
Equation of the locus of point P is x2+ y2= 18.
Question 26
Maharashtra Board Solution
Tangents to the circle x2 + y2 = a2 with inclinations, θ1 and θ2 intersect in P. Find the locus of P such that (i) tan θ1 + tan θ2 = 0 (ii) cot θ1 + cot θ2 = 5 (iii) cot θ1. cot θ2 = c
Solution & Step-by-Step Answer:
Let P(x1, y1) be a point on the required locus. Equations of the tangents to the circle x2 + y2 = a2 with slope m are y = mx ± Since, these tangents pass through (x1, y1).
