The point on the straight line , which is nearest to the circle
(1, 1)
(4, -2)
(-2, 9)
(3, -1)
Shortest distance between 2 curves occurs along the common normal. Let required point be (t, 2 -t). Normal at this will be
y –(2 –t) = (x –t), it should passes through centre of circle so
–1 –2 + t = 5 –t ⇒ t = 4, so point can be (4, -2).
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