Question

Equation of a circle which passes through (3, 6) and touches the axes is ______.

पर्याय

• x² + y² + 6x + 6y + 3 = 0

• x² + y² – 6x – 6y – 9 = 0

• x² + y² – 6x – 6y + 9 = 0

• None of these

Answer 1

Equation of a circle which passes through (3, 6) and touches the axes is x² + y² – 6x – 6y + 9 = 0.

Explanation:

Let the required circle touch the axes at (a, 0) and (0, a)

∴ Centre is (a, a) and r = a

So the equation of the circle is (x – a)² + (y – a)² = a²

If it passes through a point P(3, 6) then

(3 – a)² + (6 – a)² = a²

⇒ 9 + a² – 6a + 36 + a² – 12a = a²

⇒ a² – 18a + 45 = 0

⇒ a² – 15a – 3a + 45 = 0

⇒ a(a – 15) – 3(a – 15) = 0

⇒ (a – 3)(a – 15) = 0

⇒ a = 3 and a = 15 which is not possible

∴ a = 3

So, the required equation of the circle is (x – 3)² + (y – 3)² = 9

⇒ x² + 9 – 6x + y² + 9 – 6y = 9

⇒ x² + y² – 6x – 6y + 9 = 0