Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.1
Solution & Step-by-Step Answer:
(i) The equation of a circle with centre at origin and radius ‘r’ is given by x2 + y2 = r2 Here, r = 4 ∴ The required equation of the circle is x2 + y2 = 42 i.e., x2 + y2 = 16.
(ii) 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 = -3, k = -2 and r = 6
∴ The required equation of the circle is
[x – (-3)]2+ [y – (-2)]2= 62
⇒ (x + 3)2+ (y + 2)2= 36
⇒ x2+ 6x + 9 + y2+ 4y + 4 – 36 = 0
⇒ x2+ y2+ 6x + 4y – 23 = 0
(iii) 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 = -3 and r = 5
The required equation of the circle is
(x – 2)2+ [y – (-3)]2= 52
⇒ (x – 2)2+ (y + 3)2= 25
⇒ x2– 4x + 4 + y2+ 6y + 9 – 25 = 0
⇒ x2+ y2– 4x + 6y – 12 = 0
(iv) Centre of the circle is C (-3, -3) and it passes through the point P (-3, -6).
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 = -3, k = -3, r = 3
The required equation of the circle is
[x – (-3)]2+ [y – (-3)]2= 32
⇒ (x + 3)2+ (y + 3)2= 9
⇒ x2+ 6x + 9 + y2+ 6y + 9 – 9 = 0
⇒ x2+ y2+ 6x + 6y + 9 = 0
Check:
If the point (-3, -6) satisfies x2+ y2+ 6x + 6y + 9 = 0, then our answer is correct.
L.H.S. = x2+ y2+ 6x + 6y + 9
= (-3)2+ (-6)2+ 6(-3) – 6(-6) + 9
= 9 + 36 – 18 – 36 + 9
= 0
= R.H.S.
Thus, our answer is correct.
Solution & Step-by-Step Answer:
(i) Given equation of the circle is x2 + y2 = 25 ⇒ x2 + y2 = (5)2 Comparing this equation with x2 + y2 = r2, we get r = 5 Centre of the circle is (0, 0) and radius of the circle is 5.
(ii) Given equation of the circle is
(x – 5)2+ (y – 3)2= 20
⇒ (x – 5)2+ (y – 3)2= (√20)2
Comparing this equation with (x – h)2+ (y – k)2= r2, we get
h = 5, k = 3 and r = √20 = 2√5
Centre of the circle = (h, k) = (5, 3)
and radius of the circle = 2√5.
(iii) Given the equation of the circle is
Comparing this equation with (x – h)2+ (y – k)2= r2, we get
h = , k = and r =
Centre of the circle = (h, k) = (, ) and radius of the circle =
Solution & Step-by-Step Answer:
(i) Since the circle is touching the Y-axis, the radius of the circle is X-co-ordinate of the centre. ∴ r = a 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 = a, k = b The required equation of the circle is ⇒ (x – a)2 + (y – b)2 = a2 ⇒ x2 – 2ax + a2 + y2 – 2by + b2 = a2 ⇒ x2 + y2 – 2ax – 2by + b2 = 0
(ii) Since the circle is touching the X-axis, the radius of the circle is the Y co-ordinate of the centre.
∴ r = 3
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 = 3
The required equation of the circle is
⇒ (x + 2)2+ (y – 3)2= 32
⇒ x2+ 4x + 4 + y2– 6y + 9 = 9
⇒ x2+ y2+ 4x – 6y + 4 = 0
(iii) Let the co-ordinates of the centre of the required circle be C (h, 0).
Since the circle passes through the origin i.e., O(0, 0)
OC = radius
⇒
⇒ h2= 16
⇒ h = ±4
the co-ordinates of the centre are (4, 0) or (-4, 0).
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 = ± 4, k = 0, r = 4
The required equation of the circle is
⇒ (x – 4)2+ (y – 0)2= 42or (x + 4)2+ (y – 0)2= 42
⇒ x2– 8x + 16 + y2= 16 or x2+ 8x + 16 + y2= 16
⇒ x2+ y2– 8x = 0 or x2+ y2+ 8x = 0
(iv) Centre of the circle is C (3, 1).
Let the circle touch the line 8x – 15y + 25 = 0 at point M.
CM = radius (r)
CM = Length of perpendicular from centre C(3, 1) on the line 8x – 15y + 25 = 0
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 = 3, k = 1 and r = 2
The required equation of the circle is
⇒ (x – 3)2+ (y – 1)2= 22
⇒ x2– 6x + 9 + y2– 2y + 1 = 4
⇒ x2+ y2– 6x – 2y + 10 – 4 = 0
⇒ x2+ y2– 6x – 2y + 6 = 0
Solution & Step-by-Step Answer:
Given equations of diameters are 2x + y = 6 and 3x + 2y = 4. Let C (h, k) be the centre of the required circle. Since point of intersection of diameters is the centre of the circle, x = h, y = k Equations of diameters become 2h + k = 6 …..(i) and 3h + 2k = 4 ……..(ii) By (ii) – 2 × (i), we get -h = -8 ⇒ h = 8 Substituting h = 8 in (i), we get 2(8) + k = 6 ⇒ k = 6 – 16 ⇒ k = -10 Centre of the circle is C (8, -10) and radius, r = 9 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 = 8, k = -10 The required equation of the circle is ⇒ (x – 8)2 + (y + 10)2 = 92 ⇒ x2 – 16x + 64 + y2 + 20y + 100 = 81 ⇒ x2 + y2 – 16x + 20y + 100 + 64 – 81 = 0 ⇒ x2 + y2 – 16x + 20y + 83 = 0
Solution & Step-by-Step Answer:
y = 2x is the chord of the given circle. It satisfies the equation of a given circle. Substituting y = 2x in x2 + y2 – 10x = 0, we get ⇒ x2 + (2x)2 – 10x = 0 ⇒ x2 + 4x2 – 10x = 0 ⇒ 5x2 – 10x = 0 ⇒ 5x(x – 2) = 0 ⇒ x = 0 or x = 2 When x = 0, y = 2x = 2(0) = 0 ∴ A = (0, 0) When x = 2, y = 2x = 2 (2) = 4 ∴ B = (2, 4) End points of chord AB are A(0, 0) and B(2, 4). Chord AB is the diameter of the required circle. The equation of a circle having (x1, y1) and (x2, y2) as end points of diameter is given by (x – x1) (x – x2) + (y – y1) (y – y2) = 0 Here, x1 = 0, y1 = 0, x2 = 2, y2 = 4 The required equation of the circle is ⇒ (x – 0) (x – 2) + (y – 0) (y – 4 ) = 0 ⇒ x2 – 2x + y2 – 4y = 0 ⇒ x2 + y2 – 2x – 4y = 0
Solution & Step-by-Step Answer:
The radius of the circle = 4 units Since the circle touches both the co-ordinate axes and its centre is in the third quadrant, the centre of the circle is C(-4, -4). 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 = -4, k = -4, r = 4 the required equation of the circle is ⇒ [x – (-4)]2 + [y – (-4)]2 = 42 ⇒ (x + 4)2 + (y + 4)2 = 16 ⇒ x2 + 8x + 16 + y2 + 8y + 16 – 16 = 0 ⇒ x2 + y2 + 8x + 8y + 16 = 0
Solution & Step-by-Step Answer:
Let the circle intersect X-axis at point A and intersect Y-axis at point B. the co-ordinates of point A are (4, 0) and the co-ordinates of point B are (0, -5). Since ∠AOB is a right angle, AB represents the diameter of the circle. The equation of a circle having (x1, y1) and (x2, y2) as end points of diameter is given by (x – x1) (x – x2) + (y – y1) (y – y2) = 0 Here, x1 = 4, y1 = 0, x2 = 0, y2 = -5 The required equation of the circle is ⇒ (x – 4) (x – 0) + (y – 0) [y – (-5)] = 0 ⇒ x(x – 4) + y(y + 5) = 0 ⇒ x2 – 4x + y2 + 5y = 0 ⇒ x2 + y2 – 4x + 5y = 0
Solution & Step-by-Step Answer:
Let C(h, k) be the centre of the required circle which lies on the line x – 2y + 9 = 0. Equation of line becomes h – 2k + 9 = 0 …..(i) Also, the required circle passes through points A(1, -4) and B(5, 2). CA = CB = radius CA = CB By distance formula, Squaring both the sides, we get ⇒ (h – 1)2 + (k + 4)2 = (h – 5)2 + (k – 2)2 ⇒ h2 – 2h + 1 + k2 + 8k + 16 = h2 – 10h + 25 + k2 – 4k + 4 ⇒ -2h + 8k + 17 = -10h – 4k + 29 ⇒ 8h + 12k – 12 = 0 ⇒ 2h + 3k – 3 = 0 ……(ii) By (ii) – (i) × 2, we get 7k = 21 ⇒ k = 3 Substituting k = 3 in (i), we get h – 2(3) + 9 = 0 ⇒ h – 6 + 9 = 0 ⇒ h = -3 Centre of the circle is C(-3, 3). radius (r) = CA 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 = -3, k = 3, r = √65 The required equation of the circle is ⇒ [x – (-3)]2 + (y – 3)2 = (√65)2 ⇒ (x + 3)2 + (y – 3)2 = 65 ⇒ x2 + 6x + 9 + y2 – 6y + 9 – 65 = 0 ⇒ x2 + y2 + 6x – 6y – 47 = 0
