Question

Equation of a circle which passes through (3, 6) and touches the axes is
(a) x² + y² + 6x + 6y + 3 = 0
(b) x² + y² – 6x – 6y – 9 = 0
(c) x² + y² – 6x – 6y + 9 = 0
(d) None of these

Answer 1

Since circle touches both axis
⇒ equations of circle is of the form
(x – a)² + (y – a)² = a²
i.e. x² + a² – 2ax + y² – 2ay + a² = a²
i.e. x² + y² – 2ax – 2ay + a² = 0
Since circle passes through (3, 6)
⇒ (3)² + (6)² – 2a(3) – 2a(6) + a² = 0
i.e. 9 + 36 – 6a – 12a + a² = 0
i.e. a² – 18a + 45 = 0
i.e. a² – 15a – 3a + 45 = 0
i.e. a(a – 15) – 3(a – 15) = 0
i.e. a = 3 or a = 15
i.e. (x – 3)² + (y – 3)² = 9
i.e. x² + y² – 2 × 3x – 6y + 9 = 0
i.e. x² + y² – 6x – 6y + 9 = 0
Hence, the correct answer is option C.