Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.3
Solution & Step-by-Step Answer:
(i) Given equation of the circle is x2 + y2 = 9 ⇒ x2 + y2 = 32 Comparing this equation with x2 + y2 = r2, we get r = 3 The parametric equations of the circle in terms of θ are x = r cos θ and y = r sin θ ⇒ x = 3 cos θ and y = 3 sin θ
(ii) Given equation of the circle is
x2+ y2+ 2x – 4y – 4 = 0
⇒ x2+ 2x + y2– 4y – 4 = 0
⇒ x2+ 2x + 1 – 1 + y2– 4y + 4 – 4 – 4 = 0
⇒ (x2+ 2x + 1 ) + (y2– 4y + 4) – 9 = 0
⇒ (x + 1)2+ (y – 2)2= 9
⇒ (x + 1)2+ (y – 2)2= 32
Comparing this equation with (x – h)2+ (y – k)2= r2, we get
h = -1, k = 2 and r = 3
The parametric equations of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = -1 + 3 cos θ and y = 2 + 3 sin θ
(iii) Given equation of the circle is
(x – 3)2+ (y + 4)2= 25
⇒ (x – 3)2+ (y + 4)2= 52
Comparing this equation with (x – h)2+ (y – k)2= r2, we get
h = 3, k = -4 and r = 5
The parametric equations of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = 3 + 5 cos θ and y = -4 + 5 sin θ
Solution & Step-by-Step Answer:
Given equation of the circle is 3x2 + 3y2 – 4x + 6y – 4 = 0 Dividing throughout by 3, we get Comparing this equation with (x – h)2 + (y – k)2 = r2, we get h = , k = -1 and r = The parametric representation of the circle in terms of θ are x = h + r cos θ and y = k + r sin θ ⇒ x = + cos θ and y = -1 + sin θ
Solution & Step-by-Step Answer:
Given equation of the circle is x2 + y2 – 3x + 2y = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = -3, 2f = 2, c = 0 ⇒ g = , f = 1, 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 (0, 0) is x(0) + y(0) + () (x + 0) + 1(y + 0) + 0 = 0 ⇒ x + y = 0 ⇒ 3x – 2y = 0
Solution & Step-by-Step Answer:
Given equation of the circle is x2 + y2 + 5x – 7y + 4 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 5, 2f = -7, c = 4 ⇒ g = , f = , 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, 2) is 7x – 3y – 1 = 0, which is same as the given line. The line 7x – 3y – 1 = 0 touches the given circle at (1, 2).
Solution & Step-by-Step Answer:
Given equation of the circle is x2 + y2 – 4x + 3y + 2 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = -4, 2f = 3, c = 2 g = -2, f = , c = 2 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 (4, -2) is x(4) + y(-2) – 2(x + 4) + (y – 2) + 2 = 0 ⇒ 4x – 2y – 2x – 8 + y – 3 + 2 = 0 ⇒ 2x – y – 9 = 0 ⇒ 4x – y – 18 = 0