A circle with radius r touches the parabola y2=4x at two points such that area formed by common tangent and common chord is 32 sqaure units. the value of r is
A
√5 units
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B
2√5units
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C
3√5units
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D
4√5units
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Solution
The correct option is B2√5units
For circle to touch the parabola at two distinct points,the circle should be symmetric to the axis of the parabola as given in the figure
For given parabola y2=4x ⇒a=1
Let B(t2,2t),C(t2,−2t).So, D≡(t2,0)
Tangents at B,C intersect at A≡(−t2,0)
∴ area of required△ABC=2×12×DB×AD ⇒32=(2t)×(2t2) ⇒t=2
Now, normal at B will pass through the centre of circle
equation of Normal is y+xt=2t+3 ⇒y+2x=4+8 ⇒y+2x=12
for y=0 x=6
Hence, centre of circle is (6,0)
Radius of this circle is OB=√(6−4)2+(0−4)2=√4+16=2√5 units