Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.2
Question 1
Maharashtra Board Solution
Find the centre and radius of each of the following circles: (i) x2 + y2 – 2x + 4y – 4 = 0 (ii) x2 + y2 – 6x – 8y – 24 = 0 (iii) 4x2 + 4y2 – 24x – 8y – 24 = 0
Solution & Step-by-Step Answer:
(i) Given equation of the circle is x2 + y2 – 2x + 4y – 4 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = -2, 2f = 4 and c = -4 ⇒ g = -1, f = 2 and c = -4 Centre of the circle = (-g, -f) = (1, -2) and radius of the circle
(ii) Given equation of the circle is x2+ y2– 6x – 8y – 24 = 0
Comparing this equation with x2+ y2+ 2gx + 2fy + c = 0, we get
2g = -6, 2f = -8 and c = -24
⇒ g = -3, f = -4 and c = -24
Centre of the circle = (-g, -f) = (3, 4)
and radius of the circle
(iii) Given equation of the circle is 4x2+ 4y2– 24x – 8y – 24 = 0
Dividing throughout by 4, we get x2+ y2– 6x – 2y – 6 = 0
Comparing this equation with x2+ y2+ 2gx + 2fy + c = 0, we get
2g = -6, 2f = -2 and c = -6
⇒ g = -3, f = -1 and c = -6
Centre of the circle = (-g, -f) = (3, 1)
and radius of the circle
Question 2
Maharashtra Board Solution
Show that the equation 3x2 + 3y2 + 12x + 18y – 11 = 0 represents a circle.
Solution & Step-by-Step Answer:
Given equation is 3x2 + 3y2 + 12x + 18y – 11 = 0 Dividing throughout by 3, we get x2 + y2 + 4x + 6y – = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 4, 2f = 6, c = ⇒ g = 2, f = 3, c = Now, g2 + f2 – c = (2)2 + (3)2 – () = 4 + 9 + = > 0 ∴ The given equation represents a circle.
Question 3
Maharashtra Board Solution
Find the equation of the circle passing through the points (5, 7), (6, 6), and (2, -2).
Solution & Step-by-Step Answer:
Let C(h, k) be the centre of the required circle. Since the required circle passes through points A(5, 7), B(6, 6), and D(2, -2), CA = CB = CD = radius Consider, CA = CD By distance formula, Squaring both the sides, we get ⇒ (h – 5)2 + (k – 7)2 = (h – 2)2 + (k + 2)2 ⇒ h2 – 10h + 25 + k2 – 14k + 49 = h2 – 4h + 4 + k2 + 4k + 4 ⇒ -10h – 14k + 74 = -4h + 4k + 8 ⇒ 6h + 18k – 66 = 0 ⇒ h + 3k – 11 = 0 …..(i) Consider, CB = CD By distance formula, Squaring both the sides, we get ⇒ (h – 6)2 + (k – 6)2 = (h – 2)2 + (k + 2)2 ⇒ h2 – 12h + 36 + k2 – 12k + 36 = h2 – 4h + 4 + k2 + 4k + 4 ⇒ -12h – 12k + 72 = -4h + 4k + 8 ⇒ 8h + 16k – 64 = 0 ⇒ h + 2k – 8 = 0 ……(ii) By (i) – (ii), we get k = 3 Substituting k = 3 in (i), we get h + 3(3) – 11 = 0 ⇒ h + 9 – 11 = 0 ⇒ h = 2 Centre of the circle is C(2, 3). radius (r) = CD = = = √25 = 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 = 3 The required equation of the circle is (x – 2)2 + (y – 3)2 = 52 ⇒ x2 – 4x + 4 + y2 – 6y + 9 = 25 ⇒ x2 + y2 – 4x – 6y + 4 + 9 – 25 = 0 ⇒ x2 + y2 – 4x – 6y – 12 = 0
Question 4
Maharashtra Board Solution
Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are concyclic.
Solution & Step-by-Step Answer:
Let the equation of the circle passing through the points (3, -2), (1, 0) and (-1, -2) be x2 + y2 + 2gx + 2fy + c = 0 …..(i) For point (3, -2), Substituting x = 3 and y = -2 in (i), we get 9 + 4 + 6g – 4f + c = 0 ⇒ 6g – 4f + c = -13 ….(ii) For point (1, 0), Substituting x = 1 andy = 0 in (i), we get 1 + 0 + 2g + 0 + c = 0 ⇒ 2g + c = -1 ……(iii) For point (-1, -2), Substituting x = -1 and y = -2, we get 1 + 4 – 2g – 4f + c = 0 ⇒ 2g + 4f – c = 5 …….(iv) Adding (ii) and (iv), we get 8g = -8 ⇒ g = -1 Substituting g = -1 in (iii), we get -2 + c = -1 ⇒ c = 1 Substituting g = -1 and c = 1 in (iv), we get -2 + 4f – 1 = 5 ⇒ 4f = 8 ⇒ f = 2 Substituting g = -1, f = 2 and c = 1 in (i), we get x2 + y2 – 2x + 4y + 1 = 0 ……….(v) If (1, -4) satisfies equation (v), the four points are concyclic. Substituting x = 1, y = -4 in L.H.S of (v), we get L.H.S. = (1)2 + (-4)2 – 2(1) + 4(-4) + 1 = 1 + 16 – 2 – 16 + 1 = 0 = R.H.S. Point (1, -4) satisfies equation (v). ∴ The given points are concyclic.