CircleHard

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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