Question: Let C be a circle of radius √20 cm. Let l1 , l2 be the lines given by 2x − y −1 = 0 and x + 2y −18 = 0, respectively. Suppose that l1 passes through the center of C and that l2 is tangent to C at the point of intersection of l1 and l2 .
Explanation:
l1 : 2x – y = 12 : x + 2y = 18
Slope for l1 i.e., m1 = -2/-1 = 2 and slope for l2 i.e., m2 = -1/2
Here, m1 × m2 = -1, hence the two given lines are perpendicular
Let us first represent the figure and the 2 lines l1 and l2 .
Let O (a, b) be the centre of the circle and A be the point of intersection of the 2 lines l1 and l2 .
The coordinates of A can be found out by solving the simultaneous equations of the lines l1 (2x – y = 1) and l2 (x + 2y = 18).
Solving both these equations we get the value of x and y as 4 and 7 respectively.
∴ Coordinates of A are (4, 7).
Also, distance OA is the radius of the circle i.e., √20 units.
∴ (a - 4)2 + (b - 7)2 = 20 …(1)
Also (a, b) lies on l1 hence, 2a − b = 1
From (1) and (2) we get
(a - 4)2 + (2a - 1 - 7)2 = 20
⇒ 5a2 – 40a + 60 = 0
⇒ a2 – 8a + 12 = 0
⇒ a = 2 or 6.
If a = 2, b = 3, hence a + b = 5
If a = 6, b = 11, hence a + b = 17
∴ (a + b) is either 5 or 17.
From 5 and 17 only 17 is listed in option (c).
Hence, option (c).