JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 12)
If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x2 + y2 = 1 is a circle of radius r, then r is equal to :
$${1 \over 4}$$
$${1 \over 2}$$
1
$${1 \over 3}$$
Explanation
Let P(h, k) and point on the circle is (cos$$\theta$$, sin$$\theta$$)
$$ \therefore $$ $${{3 + \cos \theta } \over 2} = h$$ and $${{2 + \sin \theta } \over 2} = k$$
cos$$\theta$$ = 2h $$-$$ 3 and sin$$\theta$$ = 2h $$-$$ 2
Squaring and adding we get
$${(2h - 3)^2} + {(2h - 2)^2} = 1$$
$$ \Rightarrow 4{x^2} - 12x + 9 + 4{y^2} - 8y + 4 = 1$$
$$ \Rightarrow 4{x^2} + 4{y^2} - 12x - 8y + 12 = 0$$
$$ \Rightarrow {x^2} + {y^2} - 3x - 2y + 3 = 0$$
Radius = $$\sqrt {{9 \over 4} + 1 - 3} = {1 \over 2}$$
$$ \therefore $$ $${{3 + \cos \theta } \over 2} = h$$ and $${{2 + \sin \theta } \over 2} = k$$
cos$$\theta$$ = 2h $$-$$ 3 and sin$$\theta$$ = 2h $$-$$ 2
Squaring and adding we get
$${(2h - 3)^2} + {(2h - 2)^2} = 1$$
$$ \Rightarrow 4{x^2} - 12x + 9 + 4{y^2} - 8y + 4 = 1$$
$$ \Rightarrow 4{x^2} + 4{y^2} - 12x - 8y + 12 = 0$$
$$ \Rightarrow {x^2} + {y^2} - 3x - 2y + 3 = 0$$
Radius = $$\sqrt {{9 \over 4} + 1 - 3} = {1 \over 2}$$
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