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0 votes
27.0k views
in Mathematics by (21.8k points)

Equation of the circle of minimum radius which touches both the parabola y = x2 + 2x + 4 and x = y2 + 2y + 4 is

(A)  4x2 + 4y2 - 11x - 11y -31 = 0

(B)  4x2 + 4y2 - 11x - 11y -13 = 0

(A)  4x2 + 4y2 - 11x - 11y  + 11 = 0

(A)  4x2 + 4y2 - 11x - 11y - 6 = 0

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1 Answer

+1 vote
by (73.1k points)

Correct option (B)  4x2 + 4y2 - 11x - 11y -13 = 0

Explanation

The common normal to the parabola has slope – 1 and meets the parabolas at (-1/2, 13/4) , and , (13/4,-1/2). Hence, the required circle is x2 + y2- 11/4 x - 11/4 x-11/4 y - 13 /4 = 0

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