NTA Abhyas JEE Main2020MathematicsCirclePractice
The equation of the circle which passes through the points A 0,5 and B 6,1 and whose centre lies on the line 12 x + 5 y = 25 is
Options
- A3 x 2 + 3 y 2 + 10 x + 6 y + 15 = 0
- B3 x 2 + 3 y 2 - 10 x - 6 y - 45 = 0
- Cx 2 + y 2 - 6 x - 6 y + 5 = 0
- Dx 2 + y 2 - 4 x - 3 y - 10 = 0
Correct answer
B. 3 x 2 + 3 y 2 - 10 x - 6 y - 45 = 0
Step-by-step solution
Centre C of the required circle lies on the perpendicular bisector of the line segment joining 0,5 & 6,1 the midpoint of line segment joining A & B is 3,3 = D Slope of A B is - 2 3 Slope of D C is 3 2 ⇒ equation of D C is 3 x - 2 y - 3 = 0 C is the point of intersection of 3 x - 2 y - 3 = 0 and 12 x + 5 y = 25 ⇒ C = 5 3 , 1 Radius = A C = 25 9 + 16 = 13 3 ⇒ equation of the circle is x - 5 3 2 + y - 1 2 = 13 3 2 ⇒ 3 x 2 + 3 y 2 - 10 x - 6 y - 45 = 0