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प्रश्न
Equation of a circle which passes through (3, 6) and touches the axes is ______.
विकल्प
x2 + y2 + 6x + 6y + 3 = 0
x2 + y2 – 6x – 6y – 9 = 0
x2 + y2 – 6x – 6y + 9 = 0
None of these
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उत्तर
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
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