The equation of a circle which passes through the point (1, –2) and (4, –3) and whose…
Mathematics · JEE Main · NTA Exams — Co-ordinate Geometry
The equation of a circle which passes through the point (1, –2) and (4, –3) and whose centre lies on the line 3x + 4 y = 7 is
- 15 (x2 + y2) – 94 x + 18 y – 55 = 0
- 15 (x2 + y2) – 94 x + 18 y + 55 = 0
- 15 (x2 + y2) + 94 x – 18 y + 55 = 0
- none of these
Answer
(B) 15 (x 2 + y 2 ) – 94 x + 18 y + 55 = 0
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