CircleMedium
Question
The circle passing through (1, - 2) and touching the axis of x at (3, 0), also passes through the point :
Options
A.(-5, 2)
B.(2, -5)
C.(5, -2)
D.(-2, 5)
Solution
Let equation of circle be (x - 3)2 + (y + r)2 = r2
∴it passes through (1, -2)
⇒ r = 2
⇒ circle is (x - 3)2 + (y + 2)2 = 4
⇒ (5, -2)
Aliter :
(x - 3)2 + y2 + λy = 0 .........(1)
Putting (1, -2) in (1)
⇒ λ = 4
Required circle is
x2 + y2 - 6x + 4y + 9 = 0
point (5, -2) satisfies the equation.
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